Category theory

summary(technical):
A **[category](https://arbital.com/p/-4cx)** consists of a
collection of [objects](https://arbital.com/p/-object_category_theory) and a collection of [morphisms](https://arbital.com/p/-4d8) (also called arrows). To each morphism $f$ is associated a pair of objects: $\text{dom}(f)$ which is the **source** or [**domain**](https://arbital.com/p/domain_of_function) and $\text{cod}(f)$ which is the **target** or [**codomain**](https://arbital.com/p/codomain_of_function) of $f$. If $\text{dom}(f) = X$ and $\text{cod}(f) = Y$, this is written $f: X \rightarrow Y$.

These morphisms must satisfy three conditions:

 1. [**Composition**](https://arbital.com/p/Composition_of_functions): For any two morphisms $f: X \rightarrow Y$ and $g: Y \rightarrow Z$, there exists a morphism $X \rightarrow Z$, written as $g \circ f$ or simply $gf$.
 2. [**Associativity**](https://arbital.com/p/3h4): For any morphisms $f: X \rightarrow Y$, $g: Y \rightarrow Z$ and $h:Z \rightarrow W$  composition is associative, i.e., $h(gf) = (hg)f$.
 3. [**Identity**](https://arbital.com/p/identity_map): For any object $X$, there is a (unique) morphism, $1_X : X \rightarrow X$ which, when composed with another morphism, leaves it unchanged. I.e., given $f:W \rightarrow X$ and $g:X \rightarrow Y$ it holds that: $1_X f = f$ and $g 1_X = g$.

summary(motivation):
Many mathematical constructions (such as [products](https://arbital.com/p/-product_mathematics)) appear across different fields of mathematics, consisting of different ingredients but nevertheless capturing a similar idea (and often even under the same name). Category theory allows one to precisely describe the property that these different constructions all at once. This allows one to prove [theorems](https://arbital.com/p/-theorem) about all these structures at once. Hence, once you prove that a specific mathematical structure is, say, a product, then all the category-theoretic theorems about products are true for that structure. In fact, sometimes there are structures which non-obviously satisfy a category-theoretic property. Especially when [category-theoretic duality](https://arbital.com/p/duality_category_theory) is involved.

In addition, category theory allows the simple description of [functors](https://arbital.com/p/-functor), [natural transformations](https://arbital.com/p/-natural_transformation) and [adjunctions](https://arbital.com/p/-adjunction). These are mathematically powerful concepts which are very difficult to describe without the language of category theory. In fact, one of the founders of category theory, [Saunders Mac Lane](https://en.wikipedia.org/wiki/Saunders_Mac_Lane), has remarked that category theory was initially developed in order to provide a language in which to speak about natural transformations.

Powerfully, functors and adjunctions between categories allow one to translate concepts from one mathematical theory to another. They provide a "translation" (either full or partial) that allows one type of object to be viewed as another, and theorems to be translated across domains. In fact, using [duality](https://arbital.com/p/-duality_cateory_theory), very non-obvious translations can be found because a theorem in one category can be translated to its "opposite theory" in the other category. Connections which are not obvious in the language of the mathematical theories themselves, become clear in the language of category theory.

**Category theory** studies the abstraction of mathematical objects (such as [sets](https://arbital.com/p/3jz), [groups](https://arbital.com/p/3gd), and [topological spaces](https://arbital.com/p/topology)) in terms of the [morphisms](https://arbital.com/p/4d8) between them. Such a collection of objects and morphisms is a [category](https://arbital.com/p/4cx). Morphisms often represent [functions](https://arbital.com/p/3jy). For example, in the category of sets, morphisms represent all functions, in the category of groups they represent group homomorphism and in the category of topological spaces, they represent [continuous](https://arbital.com/p/continuous) maps.

Categories are usually drawn as diagrams with the objects represented by variables or points with (labeled) arrows between them representing morphisms. For this reason, morphisms are also referred to as arrows.

Morphisms do not have to represent functions. For example, any [partially ordered set](https://arbital.com/p/3rb) $(P, \leq)$ may be seen as a category where the objects are the elements of the poset and there is a (unique) morphism $x \rightarrow y$ between two elements $x$ and $y$ if and only if $x \leq y$.

## Definition ##
A **category** consists of a
collection of [objects](https://arbital.com/p/-object_category_theory) and a collection of [morphisms](https://arbital.com/p/-4d8). A morphism $f$ goes from one object, say $X$, to another, say $Y$, and is drawn as an arrow from $X$ to $Y$. Note that $X$ may equal $Y$ (in which case $f$ is referred to as an [**endomorphism**](https://arbital.com/p/endomorphism)). The object $X$ is called the **source** or [**domain**](https://arbital.com/p/domain_of_function) of $f$ and $Y$ is called the **target** or [**codomain**](https://arbital.com/p/codomain_of_function) of $f$. This is written as $f: X \rightarrow Y$.

These morphisms must satisfy three conditions:

 1. [**Composition**](https://arbital.com/p/Composition_of_functions): For any two morphisms $f: X \rightarrow Y$ and $g: Y \rightarrow Z$, there exists a morphism $X \rightarrow Z$, written as $g \circ f$ or simply $gf$.
 2. [**Associativity**](https://arbital.com/p/3h4): For any morphisms $f: X \rightarrow Y$, $g: Y \rightarrow Z$ and $h:Z \rightarrow W$  composition is associative, i.e., $h(gf) = (hg)f$.
 3. [**Identity**](https://arbital.com/p/identity_map): For any object $X$, there is a (unique) morphism, $1_X : X \rightarrow X$ which, when composed with another morphism, leaves it unchanged. I.e., given $f:W \rightarrow X$ and $g:X \rightarrow Y$ it holds that: $1_X f = f$ and $g 1_X = g$.

Note that composition is written 'backwards' since given an element $x \in X$ and two functions $f: X \rightarrow Y$ and $g: Y \rightarrow Z$, the result of applying $f$ then $g$ is $g(f(x))$ which equals $(g \circ f)(x)$.

## Motivation ##

Many mathematical constructions (such as [products](https://arbital.com/p/-product_mathematics)) appear across different fields of mathematics, consisting of different ingredients but nevertheless capturing a similar idea (and often even under the same name). Category theory allows one to precisely describe the property that these different constructions all at once. This allows one to prove [theorems](https://arbital.com/p/-theorem) about all these structures at once. Hence, once you prove that a specific mathematical structure is, say, a product, then all the category-theoretic theorems about products are true for that structure. In fact, sometimes there are structures which non-obviously satisfy a category-theoretic property. Especially when [category-theoretic duality](https://arbital.com/p/duality_category_theory) is involved.

In addition, category theory allows the simple description of [functors](https://arbital.com/p/-functor), [natural transformations](https://arbital.com/p/-natural_transformation) and [adjunctions](https://arbital.com/p/-adjunction). These are mathematically powerful concepts which are very difficult to describe without the language of category theory. In fact, one of the founders of category theory, [Saunders Mac Lane](https://en.wikipedia.org/wiki/Saunders_Mac_Lane), has remarked that category theory was initially developed in order to provide a language in which to speak about natural transformations.

Powerfully, functors and adjunctions between categories allow one to translate concepts from one mathematical theory to another. They provide a "translation" (either full or partial) that allows one type of object to be viewed as another, and theorems to be translated across domains. In fact, using [duality](https://arbital.com/p/-duality_cateory_theory), very non-obvious translations can be found because a theorem in one category can be translated to its "opposite theory" in the other category. Connections which are not obvious in the language of the mathematical theories themselves, become clear in the language of category theory.

##Categories Give an External View##
Although the objects and morphisms of a category are intended to *represent* e.g. sets and functions, from the [point of view of the category](https://arbital.com/p/-personification_of_mathematics) the objects and morphisms have no internal structure. It is not possible to talk directly about the elements of an object or how a given morphism maps elements. Instead (from the viewpoint of the category) the information about the objects and morphisms are given completely by which objects are sources and targets for the morphisms and how the morphisms are composed.

In fact, this is the strength of category theory: abstracting away the internal details allows one to focus only on relevant information and also capture information about multiple similar types of structures that act in a certain way across different mathematical theories.

This is similar to the way that a group abstracts away what elements are whilst only capturing the information of how they are 'added' or 'multiplied'.

It is also somewhat similar to the concept of a program's API (or an interface in Java); we can't see inside the program or know how it implements something, but we know what kind of inputs and outputs programs have, and what kinds of inputs and outputs a composition of such programs have.

Note that since it abstracts something away, a category does not always capture enough information for one's purposes. For example, there is addition of group homomorphisms defined pointwise. For this purpose, other structures such as [enriched categories](https://arbital.com/p/enriched_category) and [n-categories](https://arbital.com/p/n_category) may be used. However, for many purposes, categories are at a very good between isomorphic objects. This is considered a feature of category theory.

##Common Symbols: Convention##
Different texts make use of different conventions. This site makes use of the following common convention:

 - Categories are written in blackboard bold upper-case letters and are usually near the beginning of the alphabet. E.g. $\mathbb{A}, \mathbb{B}, \mathbb{C}$.
 - Objects are written as upper-case letters usually near the beginning or end of the alphabet. E.g. $A, B, C, W, X, Y, Z$.
 - Morphisms are labelled with lower-case letters, usually near f or near u. E.g. $e, f, g, h, u, v, w$.
 - Elements of an object, where necessary, are written as lower-case letters, usually near the beginning or end of the alphabet. E.g. $a, b, c, x, y, z$
 - Functors are written as upper-case letters usually near F. E.g. $E, F, G, H$.
 - Natural transformations are written as Greek letters, usually near the beginning of the alphabet. E.g. $\alpha, \beta, \gamma, \delta$.
 - The morphisms forming part of a cone or cocone for a limit or colimit are often written as Greek letters with subscripts, usually $\kappa$ or $\lambda$.

These conventions are merely guidelines and far from universally followed. Check the definition for the symbol in question to see what it represents

##[Isomorphisms in Category Theory](https://arbital.com/p/-isomorphism_category_theory)##
In category theory, [isomorphic](https://arbital.com/p/-4f4) objects are not distinguished. Many [https://arbital.com/p/-universal_constructions](https://arbital.com/p/-universal_constructions) do not pin down a [specific construction](https://arbital.com/p/-specific_construction_category_theory) but instead only specify it [up to isomorphism](https://arbital.com/p/-isomorphism_category_theory).

Doing something in category theory which relies on a specific construction, that is, requiring objects to be equal instead of merely isomorphic, is colloquially referred to as [evil](https://arbital.com/p/-evil_category_theory).

##[Universal Properties](https://arbital.com/p/-600)##
One of the most important concepts in category theory is that of a [universal property](https://arbital.com/p/600). An object in a category which satisfies a universal property is in a sense the 'best' (often meaning smallest or largest) object satisfying a certain property. This can often be used to describe in a universal way constructions like [products](https://arbital.com/p/4mj) which are defined for multiple distinct structures. In category theory, it is defined once without referring to a specific construction. This definition can then be applied to multiple categories.

The simplest [non-trivial](https://arbital.com/p/trivial) universal construction is the [terminal object](https://arbital.com/p/terminal_object). Given a category $\mathbb{C}$, an object $T$ in $\mathbb{C}$ is called a **terminal object** if, for any object $X$ in $\mathbb{C}$, there is a [unique](https://arbital.com/p/unique) morphism $f: X \rightarrow T$. In other words there is some $f: X \rightarrow T$ and if there is also $g: X \rightarrow T$ then $f=g$. In the category of sets, the terminal objects are exactly the one element sets. Given a one element set $\{a\}$, and any set $X$, there is a unique morphism $f: X \rightarrow \{a\}$, namely the function taking every $x$ in $X$ to $a$. In the category of groups, terminal objects are exactly one-element groups. Note that terminal objects need not exist. Consider a [poset seen as a category](https://arbital.com/p/poset_as_category). If it has a largest element $T$, then each object is less than or equal to $T$. So from each object there is a unique morphism to $T$ and hence it is terminal. If, however, there is no largest element then the category has no terminal object.

As another example, [products](https://arbital.com/p/4mj) can be defined by a universal property: Given a pair of objects $X$ and $Y$, an object $P$ *along with a pair of morphisms* $f: P \rightarrow X$ and $g: P \rightarrow Y$ is called the **product** of $X$ and $Y$ if, given any other object $W$ and morphisms $u: W \rightarrow X$ and $v:W \rightarrow Y$ there is a *unique* morphism $h: W \rightarrow P$ such that $fh = u$ and $gh = v$.

The above are both special cases of a very important and more general [universal construction](https://arbital.com/p/-universal_construction): the [limit](https://arbital.com/p/-limit_category_theory). This (along with the [colimit](https://arbital.com/p/-colimit_category_theory)) is described in more detail further below.

##[Duality](https://arbital.com/p/-duality_category_theory)##
For any notion in a category, its **dual** is obtained by `reversing all the arrows' and 'reversing the order of composition'. If a statement is true in any category, then its dual is true in any category. As a [corollary](https://arbital.com/p/-corollary), if a statement is true in some categories, its dual is true in the duals of those categories.

As an example, consider the definition of a [terminal object](https://arbital.com/p/-terminal_object) given above. A statement about terminal object is that any two terminal objects are isomorphic. Let's examine the exact statement. Assume $T$ is terminal. Then for any $X$ there is unique $f: X \rightarrow T$. If we reverse the arrows, we get that for every $X$ there is unique $f: X \leftarrow T$. This is the definition of an [initial object](https://arbital.com/p/-initial_object). Consider another terminal object $T'$. The statement that $T'$ is isomorphic to $T$ is means that there is some $f: T \rightarrow T'$ and $g: T' \rightarrow T$ such that $gf = 1_T$ and $fg = 1_{T'}$. The dual of this is just the statement that there is some $f: T \leftarrow T'$ and $g: T' \leftarrow T$ such that $fg = 1_T$ and $gf = 1_{T'}$, this is exactly the same property! (The morphisms $f$ and $g$ have just been renamed). Hence, the dual of the statement that a terminal object is unique up to isomorphism is the statement that every initial object is unique up to isomorphism.

Similarly, if something is true for every category with an initial object, its dual will be true for every category with a terminal object.

The concept of duality can be a powerful way of obtaining new results which come easily within category theory, but which are not obvious in the theory to which category theory is being applied. As an advanced example, the category of [Boolean Algebras](https://arbital.com/p/-boolean_algebra) is dual to the category of [Stone Spaces](https://arbital.com/p/-stone_space). See, [Stone Duality on Wikipedia](https://en.wikipedia.org/wiki/Stone's_representation_theorem_for_Boolean_algebras) for the motivation.

##[Functors](https://arbital.com/p/-Functor)##
A **functor** is a morphism between categories.

Given two categories $\mathbb{A}$ and $\mathbb{B}$, a functor $F$ from $\mathbb{A}$ to $\mathbb{B}$, written $F: \mathbb{A} \rightarrow \mathbb{B}$ is defined as a pair of functions:

 - $F_0:$ Objects($\mathbb{A}$) $\rightarrow$ Objects($\mathbb{B}$)
 - $F_1:$ Morphisms($\mathbb{A}$) $\rightarrow$ Morphisms($\mathbb{B}$)

Which satisfy:

 1.
Preservation of domain and codomain:
If $f: X \rightarrow Y$ then $F_1(f): F_0(X) \rightarrow F_1(Y)$. (
Put differently,
Dom($F_1(f)$) = $F_0$(Dom($f$)) and Cod($F_1(f)$) = $F_0$(Cod($f$)) for every morphism $f$. )

 2. Preservation of Identity:
If the morphism $1_X: X \rightarrow X$ is the identity on $X$, then the morphism $F_1(1_X): F_0(X) \rightarrow F_0(X)$ is the identity on $F_0(X)$.

 3. Preservation of composition:
Given morphisms $f: X \rightarrow Y$ and $g: Y \rightarrow Z$, then the composition of their images $F_1(g) \circ F_1(f): F_0(X) \rightarrow F_0(Z)$ is the image of their composition $F_1(g \circ f): F_0(X) \rightarrow F_0(Z)$.

Instead of differentiating $F_0$ and $F_1$, they are usually both written simply as $F$. E.g. $F(f): F(X) \rightarrow F(Y)$.

##[Properties of Morphisms](https://arbital.com/p/-morphism_category_theory)##
Morphisms are the central objects of study in category theory. For this reason, properties of morphisms can be very important.
A morphism $f: X \rightarrow Y$ is called the following if it satisfies the given property:

 - **[Isomorphism](https://arbital.com/p/-isomorphism_category_theory):**
 ([Self-dual](https://arbital.com/p/-self_dual_category_theory))

There exists some $g: Y \rightarrow X$ such that $gf = 1_X$ and $fg = 1_Y$.

Intuitively, an isomorphism is a way of transforming from one object to another in a way that makes them indistinguishable using the information of the category.

 - **[Monomorphism](https://arbital.com/p/-monomorphism):** ([Dual](https://arbital.com/p/-duality_category_theory) to epimorphic)

For any object $W$ and morphisms $g,h: W \rightarrow X$, if $fg = fh$ then $g = h$.

Intuitively, $f$ being a monomorphism indicates that *all of the information captured by the collection of morphisms into $X$ is preserved when composing by $f$*. It generalizes the notion of an [injective function](https://arbital.com/p/4b7), since in most [concrete categories](https://arbital.com/p/-concrete_category) (like [sets](https://arbital.com/p/-3jz), [groups](https://arbital.com/p/-3gd), and [topological spaces](https://arbital.com/p/-topology)) every injective map is a monomorphism. However, even in concrete categories (and certainly more generally), monomorphisms need not be injective.

 - **[Epimorphism](https://arbital.com/p/-epimorphism):** (Dual to monomorphic)

For any object $Z$ and morphisms $g,h: X \rightarrow Z$, if $gf = hf$ then $g = h$.

Intuitively, $f$ being an epimorphism indicates that *all the information captured by the collection of morphisms out of $Y$ is preserved when composing by $f$.*. It generalizes the notion of a [surjective function](https://arbital.com/p/-4bg). However, in an even stronger sense than for monomorphisms, a function being epimorphic and a function being surjective are far from equivalent.

Properties that more closely match surjectivity include [Section / Split Epimorphism](https://arbital.com/p/-section_category_theory), and [regular epimorphism](https://arbital.com/p/-regular_epimorphism). [strict epimorphism](https://arbital.com/p/-strict_epimorphism), [strong epimorphism](https://arbital.com/p/-strong_epimorphism), and [extremal epimorphism](https://arbital.com/p/-extremal_epimorphism). Note that despite the names, not all of these are necessarily epimorphisms, but are epimorphisms in "nice" categories.

 - **[Endomorphism](https://arbital.com/p/-endomorphism):** (Self-dual)

$X = Y$, i.e., $f: X \rightarrow X$.

An endomorphism is a morphism from an object to itself.

 - **[Automorphism](https://arbital.com/p/-automorphism):** (Self-dual)

The morphism $f$ is both an endomorphism and an isomorphism.

An automorphism is a morphism from a structure to itself that preserves all the information of the structure that is distinguishable by the category. Intuitively, it gives "another view" of a structure (say, by moving around its elements) that leaves it essentially unchanged.

 - **[Retraction / Split Monomorphism](https://arbital.com/p/-retraction_category_theory):** (Dual to split epimorphic)

There exists some $g: Y \rightarrow X$ such that $gf = 1_X$

A morphism is a **retraction** if its effect can be "reversed" or inverted by another morphism applied after it. For example, every injective map is a retraction. The morphism which inverts the retraction is a **section**.

 - **[Section / Split Epimorphism](https://arbital.com/p/-section_category_theory):** (Dual to split monomorphic)

There exists some $g: Y \rightarrow X$ such that $fg = 1_Y$

A morphism is a **section** if it "reverses" the effect of some other morphism applied before it. The morphism which is inverted is a **retraction**.

##[Limits](https://arbital.com/p/-limit_category_theory) and [Colimits](https://arbital.com/p/-colimit_category_theory)##

##[Further Universal Constructions](https://arbital.com/p/-universal_construction)##

##[Natural Transformations](https://arbital.com/p/-natural_transformation)##

##[Constructions on Categories](https://arbital.com/p/-Constructions_on_categories)##

##[Adjunctions and Adjoint Functors](https://arbital.com/p/-adjunction_category_theory)##

Tsvi BT
Jaime Sevilla Molina
Daniel Satanove
Mark Chimes
Eric Bruylant
Nate Soares
Patrick Stevens

The composition of two group homomorphisms is a homomorphism


Log base infinity

There is no $\log_{\infty},$ because $\infty$ is not a [https://arbital.com/p/-4bc](https://arbital.com/p/-4bc). Nevertheless, the function $z$ defined as $z(x) = 0$ for all $x \in$ [$\mathbb R^+$](https://arbital.com/p/positive_reals) can pretty readily be interpreted as $\log_{\infty}$.

That is, $z$ satisfies all properties of the basic [properties of the logarithm](https://arbital.com/p/4bz) except for the one that says there exists a $b$ such that $\log(b) = 1.$ In the case of $\log_\infty,$ the logarithm base infinity claims "well, if you gave me a $b$ that was _large enough_ I might return 1, but for all measly finite numbers I return 0." In fact, if you're feeling ambitious, you can define $\log_\infty$ to be a [https://arbital.com/p/-multifunction](https://arbital.com/p/-multifunction) which allows infinite inputs, and define $\log_\infty(\infty)$ to return any positive real number that you'd like (1 included). This requires a few hijinks (like defining $\infty^0$ to also return any number that you'd like), but can be made to work and satisfy all the basic logarithm properties (if you strategically re-interpret some '$=$' signs as '[$\in$](https://arbital.com/p/set_contains)' signs).

The moral of the story is that functions that send everything to zero are _almost_ logarithm functions, with the minor caveat that they utterly destroy all the intricate structure that logarithm functions tap into. (That's what happens when you choose "0" as your arbitrary scaling factor when tapping into [https://arbital.com/p/-4gp](https://arbital.com/p/-4gp).)


Two independent events

$$
\newcommand{\bP}{\mathbb{P}}
$$

summary:
$$
\newcommand{\bP}{\mathbb{P}}
$$
We say that two [events](https://arbital.com/p/event_probability), $A$ and $B$, are *independent* when learning that $A$ has occurred does not change your probability that $B$ occurs. That is, $\bP(B \mid A) = \bP(B)$.
Another way to state independence is that $\bP(A,B) = \bP(A) \bP(B)$.

We say that two [events](https://arbital.com/p/event_probability), $A$ and $B$, are *independent* when learning that $A$ has occurred does not change your probability that $B$ occurs. That is, $\bP(B \mid A) = \bP(B)$.
Equivalently,  $A$ and $B$ are independent if $\bP(A)$ doesn't change if you condition on $B$: $\bP(A \mid B) = \bP(A)$.

Another way to state independence is that $\bP(A,B) = \bP(A) \bP(B)$.

All these definitions are equivalent:

$$\bP(A,B) = \bP(A)\;  \bP(B \mid A)$$

by the [chain rule](https://arbital.com/p/chain_rule_probability), so

$$\bP(A,B) = \bP(A)\;  \bP(B)\;\; \Leftrightarrow \;\; \bP(A)\;  \bP(B \mid A) = \bP(A)\;  \bP(B) \ ,$$

and similarly for $\bP(B)\;  \bP(A \mid B)$.


Cycle type of a permutation

Given an element $\sigma$ of a [https://arbital.com/p/-497](https://arbital.com/p/-497) $S_n$ on finitely many elements, we may express $\sigma$ in [cycle notation](https://arbital.com/p/49f).
The cycle type of $\sigma$ is then a list of the lengths of the cycles in $\sigma$, where conventionally we omit length-$1$ cycles from the cycle type.
Conventionally we list the lengths in decreasing order, and the list is presented as a comma-separated collection of values.

The concept is well-defined because [https://arbital.com/p/-49k](https://arbital.com/p/-49k) up to reordering of the cycles.

# Examples

- The cycle type of the element $(123)(45)$ in $S_7$ is $3,2$, or (without the conventional omission of the cycles $(6)$ and $(7)$) $3,2,1,1$.
- The cycle type of the identity element is the empty list.
- The cycle type of a $k$-cycle is $k$, the list containing a single element $k$.


Two independent events: Square visualization

$$
\newcommand{\true}{\text{True}}
\newcommand{\false}{\text{False}}
\newcommand{\bP}{\mathbb{P}}
$$

summary:
$$
\newcommand{\true}{\text{True}}
\newcommand{\false}{\text{False}}
\newcommand{\bP}{\mathbb{P}}
$$

Say $A$ and $B$ are independent [events](https://arbital.com/p/event_probability), so $\bP(A, B) = \bP(A)\bP(B).$ Then we can draw their joint probability distribution using the using the [square visualization](https://arbital.com/p/496) of probabilities:

This is what independence looks like, using the [square visualization](https://arbital.com/p/496) of probabilities:

We can see that the [events](https://arbital.com/p/event_probability) $A$ and $B$ don't interact; we say that $A$ and $B$ are *independent*. Whether we look at the whole square, or just the red part of
the square where $A$ is true, the probability of $B$ stays the same. In other words, $\bP(B \mid A) = \bP(B)$. That's what we mean by independence: the
probability of $B$ doesn't change if you condition on $A$.

Our square of probabilities can be generated by multiplying together the probability of $A$ and the probability of $B$:

This picture demonstrates another way to define what it means for $A$ and $B$ to be independent:

$$\bP(A, B) = \bP(A)\bP(B)\ .$$

In terms of factoring a joint distribution
--

Let's contrast independence with non-independence. Here's a picture of two ordinary, non-independent events $A$ and $B$:

(If the meaning of this picture isn't clear, take a look at [https://arbital.com/p/496](https://arbital.com/p/496).)

We have the red blocks for $\bP(A)$ and the blue blocks for $\bP(\neg A)$ lined up in columns. This means we've [factored](https://arbital.com/p/factoring_probability) our
probability distribution using $A$ as the first factor:

$$\bP(A,B) = \bP(A) \bP(B \mid A)\ .$$

We could just as well have factored by $B$ first: $\bP(A,B) = \bP(B) \bP( A \mid B)\ .$ Then we'd draw a picture like this:

Now, here again is the picture of [two independent events](https://arbital.com/p/4cf) $A$ and $B$:

In this picture, there's red and blue lined-up columns for $\bP(A)$ and $\bP(\neg A)$, and there's *also*  dark and light lined-up rows for $\bP(B)$ and
$\bP(\neg B)$. It looks like we somehow [factored](https://arbital.com/p/factoring_probability) our probability distribution $\bP$ using both $A$ and
$B$ as the first factor.

In fact, this is exactly what happened: since $A$ and $B$ are [independent](https://arbital.com/p/4cf), we have that $\bP(B \mid A) = \bP(B)$. So the diagram
above is actually factored according to $A$ first: $\bP(A,B) = \bP(A) \bP(B \mid A)$. It's just that $\bP(B \mid A)= \bP(B) = \bP(B \mid \neg A)$, since $B$
is independent from $A$. So we don't need to have different ratios of dark to light (a.k.a. conditional probabilities of $B$) in the left and right columns:

In this visualization, we can see what happens to the probability of $B$ when you condition on $A$ or on $\neg A$: it doesn't change at all. The ratio of
\[area where $B$ happens\](https://arbital.com/p/the) to \[whole area\](https://arbital.com/p/the), is the same as the ratio $\bP(B \mid A)$ where we only look at the area where $A$ happens, which is the
same as the ratio $\bP(B \mid \neg A)$ where we only look at the area where $\neg A$ happens. The fact that the probability of $B$ doesn't change when we
condition on $A$ is exactly what we mean when we say that $A$ and $B$ are independent.

The square diagram above is *also* factored according to $B$ first, using $\bP(A,B) = \bP(B) \bP(A \mid B)$. The red / blue ratios are the same in both rows
because $\bP(A \mid B) = \bP(A) = \bP(A \mid \neg B)$, since $A$ and $B$ are independent:

We couldn't do any of this stuff if the columns and rows didn't both line up. (Which is good, because then we'd have proved the false statement that any two
events are independent!)

In terms of multiplying marginal probabilities
---

Another way to say that $A$ and $B$ are independent variables %note:We're using  the [equivalence](https://arbital.com/p/event_variable_equivalence) between [https://arbital.com/p/event_probability
events](https://arbital.com/p/event_probability
events) and [binary variables](https://arbital.com/p/binary_random_variable).% is that for any truth values $t_A,t_B \in \{\true, \false\},$

$$\bP(A = t_A, B= t_B) = \bP(A = t_A)\bP(B = t_B)\ .$$

So the [joint probabilities](https://arbital.com/p/1rh) for $A$ and $B$ are computed by separately getting the probability of $A$ and the probability of $B$, and then
multiplying the two probabilities together. For example, say we want to compute the probability $\bP(A, \neg B) = \bP(A = \true, B = \false)$. We start with
the [marginal probability](https://arbital.com/p/marginal_probability) of $A$:

and the probability of $\neg B$:

and then we multiply them:

We can get all the joint probabilities this way. So we can visualize the whole joint distribution as the thing that you get when you multiply two independent
probability distributions together. We just overlay the two distributions:

To be a little more mathematically elegant, we'd use the [topological product of two spaces](https://arbital.com/p/topological_product) shown earlier to draw the joint distribution
as a product of the distributions of $A$ and $B$:

Tsvi BT
Eric Rogstad
Jaime Sevilla Molina


Transposition (as an element of a symmetric group)

In a [https://arbital.com/p/-497](https://arbital.com/p/-497), a transposition is a permutation which has the effect of swapping two elements while leaving everything else unchanged.
More formally, it is a permutation of [order](https://arbital.com/p/order_of_a_group_element) $2$ which fixes all but two elements.

%%%knows-requisite([https://arbital.com/p/4cg](https://arbital.com/p/4cg)):
A transposition is precisely an element with cycle type $2$.
%%%

# Example

In $S_5$, the permutation $(12)$ is a transposition: it swaps $1$ and $2$ while leaving all three of the elements $3,4,5$ unchanged.
However, the permutation $(124)$ is not a transposition, because it has order $3$, not order $2$.


Every member of a symmetric group on finitely many elements is a product of transpositions

Given a permutation $\sigma$ in the [https://arbital.com/p/-497](https://arbital.com/p/-497) $S_n$, there is a finite sequence $\tau_1, \dots, \tau_k$ of [transpositions](https://arbital.com/p/4cn) such that $\sigma = \tau_k \tau_{k-1} \dots \tau_1$.
Equivalently, symmetric groups are generated by their transpositions.

Note that the transpositions might "overlap".
For example, $(123)$ is equal to $(23)(13)$, where the element $3$ appears in two of the transpositions.

Note also that the sequence of transpositions is by no means uniquely determined by $\sigma$.

# Proof

It is enough to show that a [cycle](https://arbital.com/p/49f) is expressible as a sequence of transpositions.
Once we have this result, we may simply replace the successive cycles in $\sigma$'s disjoint cycle notation by the corresponding sequences of transpositions, to obtain a longer sequence of transpositions which multiplies out to give $\sigma$.

It is easy to verify that the cycle $(a_1 a_2 \dots a_r)$ is equal to $(a_{r-1} a_r) (a_{r-2} a_r) \dots (a_2 a_r) (a_1 a_r)$.
Indeed, that product of transpositions certainly does not move anything that isn't some $a_i$; while if we ask it to evaluate $a_i$, then the $(a_1 a_r)$ does nothing to it, $(a_2 a_r)$ does nothing to it, and so on up to $(a_{i-1} a_r)$.
Then $(a_i a_r)$ sends it to $a_r$; then $(a_{i+1} a_r)$ sends the resulting $a_r$ to $a_{i+1}$; then all subsequent transpositions $(a_{i+2} a_r), \dots, (a_{r-1} a_r)$ do nothing to the resulting $a_{i+1}$.
So the output when given $a_i$ is $a_{i+1}$.

# Why is this useful?

It can make arguments simpler: if we can show that some property holds for transpositions and that it is closed under products, then it must hold for the entire symmetric group.


Order of a group element

Given an element $g$ of group $(G, +)$ (which henceforth we abbreviate simply as $G$), the order of $g$ is the number of times we must add $g$ to itself to obtain the identity element $e$.

%%%knows-requisite([https://arbital.com/p/3gg](https://arbital.com/p/3gg)):
Equivalently, it is the order of the group $\langle g \rangle$ generated by $g$: that is, the order of $\{ e, g, g^2, \dots, g^{-1}, g^{-2}, \dots \}$ under the inherited group operation $+$.
%%%

Conventionally, the identity element itself has order $1$.

# Examples

%%%knows-requisite([https://arbital.com/p/497](https://arbital.com/p/497)):
In the [https://arbital.com/p/-497](https://arbital.com/p/-497) $S_5$, the order of an element is the [https://arbital.com/p/-least_common_multiple](https://arbital.com/p/-least_common_multiple) of its [cycle type](https://arbital.com/p/4cg).
%%%
%%%knows-requisite([https://arbital.com/p/47y](https://arbital.com/p/47y)):
In the [https://arbital.com/p/-47y](https://arbital.com/p/-47y) $C_6$, the order of the generator is $6$.
If we view $C_6$ as being the integers [modulo](https://arbital.com/p/modular_arithmetic) $6$ under addition, then the element $0$ has order $1$; the elements $1$ and $5$ have order $6$; the elements $2$ and $4$ have order $3$; and the element $3$ has order $2$.
%%%

In the group $\mathbb{Z}$ of [integers](https://arbital.com/p/48l) under addition, every element except $0$ has infinite order. $0$ itself has order $1$, being the identity.


Strong Church Turing thesis

> Every realistic model of computation is [polynomial time reducible](https://arbital.com/p/) to [probabilistic Turing machines](https://arbital.com/p/)

Which amounts to saying that every computable process in the universe can be efficiently simulated by a probabilistic Turing machine.

The definition of *realistic model* appeals to intuition rather than a precise definition of realistic. A rule of thumb is that a computation model is realistic if it could be used to accurately model some physical process. For example, there is a clear relation between the model of [register machines](https://arbital.com/p/) and the inner workings of personal computers. On the other hand, a computational model which can access an [NP oracle](https://arbital.com/p/) does not have a physical counterpart.

As it happened with the standard [https://arbital.com/p/4bw](https://arbital.com/p/4bw), this is an inductive rather than a mathematically defined statement. However, unlike the standard CT thesis, it is [widely believed to be wrong](https://arbital.com/p/), primarily because [quantum computation](https://arbital.com/p/) stands as a highly likely counterexample to the thesis.

We remark that non-deterministic computation is not a candidate counterexample to the strong CT thesis, since it is not realistic. That said, we can find a relation between the two concepts: if [$P=NP$](https://arbital.com/p/4bd) and [$BQP\subset NP$](https://arbital.com/p/) then $BQP = P$, so quantum computation can be efficiently simulated, and we lose the strongest contestant for a counterexample to Strong CT.

#Consequences of falsehood

The falsehood of the Strong CT thesis opens up the possibility of the existence of physical processes that, while computable, cannot be modeled in a reasonable time with a classical computer. One consequence of this, that would also mean that if quantum computing is possible the speed .up it provides is non-trivial.

Epistemic processes which assign priors using a penalty for computational time complexity are misguided if the Strong CT thesis is false. See for example [Levin search](https://arbital.com/p/).

 [probably not worth raising? human brains have crazy-low decoherence times.](https://arbital.com/p/comment:)

Eric Bruylant
Eric Rogstad
Patrick Stevens
Jaime Sevilla Molina


Category (mathematics)

A **category** consists of a collection of objects with morphisms between them. A morphism $f$ goes from one object, say $X$, to another, say $Y$, and is drawn as an arrow from $X$ to $Y$. Note that $X$ may equal $Y$ (in which case $f$ is referred to as an [https://arbital.com/p/-endomorphism](https://arbital.com/p/-endomorphism)). The object $X$ is called the source or [domain](https://arbital.com/p/3js) of $f$ and $Y$ is called the target or [codomain](https://arbital.com/p/3lg) of $f$, though note that $f$ itself need not be a [https://arbital.com/p/-3jy](https://arbital.com/p/-3jy) and $X$ and $Y$ need not be sets. This is written as $f: X \rightarrow Y$.

These morphisms must satisfy three conditions:

 1. [**Composition**](https://arbital.com/p/Composition_of_functions): For any two morphisms $f: X \rightarrow Y$ and $g: Y \rightarrow Z$, there exists a morphism $X \rightarrow Z$, written as $g \circ f$ or simply $gf$.
 2. [**Associativity**](https://arbital.com/p/3h4): For any morphisms $f: X \rightarrow Y$, $g: Y \rightarrow Z$ and $h:Z \rightarrow W$  composition is associative, i.e., $h(gf) = (hg)f$.
 3. [**Identity**](https://arbital.com/p/identity_map): For any object $X$, there is a (unique) morphism, $1_X : X \rightarrow X$ which, when composed with another morphism, leaves it unchanged. I.e., given $f:W \rightarrow X$ and $g:X \rightarrow Y$ it holds that: $1_X f = f$ and $g 1_X = g$.

Eric Bruylant
Mark Chimes
Patrick Stevens


Dihedral group

The dihedral group $D_{2n}$ is the group of symmetries of the $n$-vertex [https://arbital.com/p/-regular_polygon](https://arbital.com/p/-regular_polygon).

# Presentation
The dihedral groups have very simple [presentations](https://arbital.com/p/group_presentation): $$D_{2n} \cong \langle a, b \mid a^n, b^2, b a b^{-1} = a^{-1} \rangle$$
The element $a$ represents a rotation, and the element $b$ represents a reflection in any fixed axis.

# Properties

- The dihedral groups $D_{2n}$ are all non-abelian for $n > 2$. ([Proof.](https://arbital.com/p/4d0))
- The dihedral group $D_{2n}$ is a [https://arbital.com/p/-subgroup](https://arbital.com/p/-subgroup) of the [https://arbital.com/p/-497](https://arbital.com/p/-497) $S_n$, generated by the elements $a = (123 \dots n)$ and $b = (2, n)(3, n-1) \dots (\frac{n}{2}+1, \frac{n}{2}+3)$ if $n$ is even, $b = (2, n)(3, n-1)\dots(\frac{n-1}{2}, \frac{n+1}{2})$ if $n$ is odd.

# Examples

## $D_6$, the group of symmetries of the triangle

# Infinite dihedral group

The infinite dihedral group has presentation $\langle a, b \mid b^2, b a b^{-1} = a^{-1} \rangle$.
It is the "infinite-sided" version of the finite $D_{2n}$.

We may view the infinite dihedral group as being the subgroup of the group of [homeomorphisms](https://arbital.com/p/homeomorphism) of $\mathbb{R}^2$ generated by a reflection in the line $x=0$ and a translation to the right by one unit.
The translation is playing the role of a rotation in the finite $D_{2n}$.


Dihedral groups are non-abelian

Let $n \geq 3$. Then the [https://arbital.com/p/-4cy](https://arbital.com/p/-4cy) on $n$ vertices, $D_{2n}$, is not [abelian](https://arbital.com/p/3h2).

# Proof

The most natural dihedral group [presentation](https://arbital.com/p/group_presentation) is $\langle a, b \mid a^n, b^2, bab^{-1} = a^{-1} \rangle$.
In particular, $ba = a^{-1} b = a^{-2} a b$, so $ab = ba$ if and only if $a^2$ is the identity.
But $a$ is the rotation which has order $n > 2$, so $ab$ cannot be equal to $ba$.


Needs image

Meta tag for pages which would benefit from having images.

If a specific image has been described, use the [https://arbital.com/p/5v6](https://arbital.com/p/5v6) tag.

Alexei Andreev
Mark Chimes
Stephanie Zolayvar
Eric Bruylant
M Yass
Patrick Stevens


Morphism

A morphism is the abstract representation of a [relation](https://arbital.com/p/3nt) between [mathematical objects](https://arbital.com/p/-mathematical_object).

Usually, it is used to refer to [functions](https://arbital.com/p/3jy) mapping element of one set to another, but it may represent a more general notion of a relation in [https://arbital.com/p/-4c7](https://arbital.com/p/-4c7).

##Isomorphisms##

To understand a morphism, it is easier to first understand the concept of an [https://arbital.com/p/-4f4](https://arbital.com/p/-4f4). Two mathematical structures (say two [groups](https://arbital.com/p/-3gd)) are called **isomorphic** if they are indistinguishable using the information of the language and theory under consideration.

Imagine you are the Count von Count. You care only about counting things. You don't care what it is you count, you just care how many there are. You decide that you want to collect objects you count into boxes, and you consider two boxes equal if there are the same number of elements in both boxes. How do you know if two boxes have the same number of elements? You pair them up and see if there are any left over in either box. If there aren't any left over, then the boxes are "bijective" and the way that you paired them up is a [bijection](https://arbital.com/p/499). A bijection is a simple form of an isomorphism and the boxes are said to be isomorphic.

For example, the theory of groups only talks about the way that elements are combined via group operation (and whether they are the [https://arbital.com/p/-identity](https://arbital.com/p/-identity) or [inverses](https://arbital.com/p/-4sn), but that information is already given by the information of how elements are combined under the group operation (hereafter called multiplication). The theory does not care in what order elements are put, or what they are labelled or even what they are. Hence, if you are using the language and theory of groups, you want to say two groups are essentially indistinguishable if their multiplication acts the same way.

Jaime Sevilla Molina
Mark Chimes
Eric Bruylant
Team Arbital
Patrick Stevens


Isomorphism

A pair of mathematical structures are **isomorphic** to each other if they are "essentially the same", even if they aren't necessarily equal.

An **isomorphism** is a [https://arbital.com/p/-4d8](https://arbital.com/p/-4d8) between isomorphic structures which translates one to the other in a way that preserves all the relevant structure. An important property of an isomorphism is that it can be 'undone' by its [inverse](https://arbital.com/p/-4sn) isomorphism.

An isomorphism from an object to itself is called an **[automorphism](https://arbital.com/p/automorphism)**. They can be thought of as symmetries: different ways in which an object can be mapped onto itself without changing it.

##Equality and Identity##
The simplest isomorphism is equality: if two things are equal then they are actually the same thing (and so not actually *two* things at all). Anything is obviously indistinguishable from itself under whatever measure you might use (it has any property in common with itself) and so regardless of the theory or language, anything is isomorphic to itself. This is represented by the [identity](https://arbital.com/p/-identity_function) (iso)morphism.

%%%knows-requisite([https://arbital.com/p/3gd](https://arbital.com/p/3gd)):
##[Group Isomorphisms](https://arbital.com/p/49x)##
For a more technical example, the theory of groups only talks about the way that elements are combined via group operation. The theory does not care in what order elements are put, or what they are labelled or even what they are. Hence, if you are using the language and theory of groups, you want to say two groups are essentially indistinguishable if you can pair up the elements such that their group operations act the same way.
%%%

##Isomorphisms in Category Theory##
In [category theory](https://arbital.com/p/-4c7), an isomorphism is a morphism which has a two-sided [https://arbital.com/p/-4sn](https://arbital.com/p/-4sn). That is to say, $f:A \to B$ is an isomorphism if there is a morphism $g: B \to A$ where $f$ and $g$ cancel each other out.

Formally, this means that both composites $fg$ and $gf$ are equal to identity morphisms (morphisms which 'do nothing' or declare an object equal to itself). That is, $gf = \mathrm {id}_A$ and $fg = \mathrm {id}_B$.

Daniel Satanove
Eric Rogstad
Mark Chimes
Eric Bruylant
Dylan Hendrickson
Patrick Stevens


Square visualization of probabilities on two events: (example) Diseasitis

$$
\newcommand{\bP}{\mathbb{P}}
$$

summary: $$ \newcommand{\bP}{\mathbb{P}} $$

We can visualize the [https://arbital.com/p/22s](https://arbital.com/p/22s):

Then we can see at a glance that the probability $\bP(D \mid B)$ of the patient having Diseasitis given that the tongue depressor is black, despite some intuitions, isn't all that big:

$$\bP(D \mid B) = \frac{\bP(B, D)}{ \bP(B, D) + \bP(B, \neg D)} = \frac{3}{7} $$

From the [https://arbital.com/p/22s](https://arbital.com/p/22s):

> You are screening a set of patients for a disease, which we'll call Diseasitis. Based on prior epidemiology, you expect that around 20% of the patients in the screening population will in fact have Diseasitis. You are testing for the presence of the disease using a tongue depressor containing a chemical strip. Among patients with Diseasitis, 90% turn the tongue depressor black. However, 30% of the patients without Diseasitis will also turn the tongue depressor black. Among all the patients with black tongue depressors, how many have Diseasitis?

It seems like, since Diseasitis very strongly predicts that the patient has a black tongue depressor, it should be the case that the [https://arbital.com/p/-1rj](https://arbital.com/p/-1rj) $\bP( \text{Diseasitis} \mid \text{black tongue depressor})$ is big. But actually, it turns out that a patient with a black tongue depressor is more likely than not to be completely Diseasitis-free.

Can we see this fact at a glance? Below, we'll use the [https://arbital.com/p/-496](https://arbital.com/p/-496) to draw pictures and use our visual intuition.

To introduce some notation: our [prior probability](https://arbital.com/p/1rm) $\bP(D)$ that the patient has Diseasitis is $0.2$. We think that if the patient is sick $(D)$, then it's 90% likely that the tongue depressor will turn black $(B)$: we assign conditional probability $\bP(B \mid D) = 0.9$. We assign conditional probability $\bP(B \mid \neg D) = 0.3$ that the tongue depressor will be black even if the patient isn't sick. We want to know $\bP(D \mid B)$, the [posterior probability](https://arbital.com/p/1rp) that the patient has Diseasitis given that we've seen a black tongue depressor.

If we wanted to, we could solve this problem precisely using [https://arbital.com/p/1lz](https://arbital.com/p/1lz):

$$
\begin{align}
\bP(D \mid B) &= \frac{\bP(B \mid D) \bP(D)}{\bP(B)}\\
&= \frac{0.9 \times 0.2}{ \bP(B, D) + \bP(B, \neg D)}\\
&= \frac{0.18}{ \bP(D)\bP(B \mid D) + \bP(\neg D)\bP(B \mid \neg D)}\\
&= \frac{0.18}{ 0.18 + 0.24}\\
&= \frac{0.18}{ 0.42} = \frac{3}{7}  \approx 0.43\ .
\end{align}
$$

So even if we've seen a black tongue depressor, the patient is more likely to be healthy than not: $\bP(D \mid B) < \bP(\neg D \mid B) \approx 0.57$.

Now, this calculation might be enlightening if you are a real expert at [https://arbital.com/p/1lz](https://arbital.com/p/1lz). A better calculation would probably be the [odds ratio form of Bayes's rule](https://arbital.com/p/1x9).

But either way, maybe there's still an intuition saying that, come on, if the tongue depressor is such a strong indicator of Diseasitis that $\bP(B \mid D) = 0.9$, it must be that $\bP(D \mid B) =big$.

Let's use the [square visualization of probabilities](https://arbital.com/p/496) to make it really visibly obvious that $\bP(D \mid B) < \bP(\neg D \mid B)$, and to figure out why $\bP(B \mid D) = big$ doesn't imply $\bP(D \mid B) =big$.

We start with the probability of $\bP(D)$ (so we're  [factoring](https://arbital.com/p/factoring_probability) our probabilities by $D$ first):

Now let's break up the red column, where $D$ is true and the patient has Diseasitis, into a block for the probability $\bP(B \mid D)$ that $B$ is also true, and a block for the probability $\bP(\neg B \mid D)$ that $B$ is false.

>Among patients with Diseasitis, 90% turn the tongue depressor black.

That is, in 90% of the  outcomes where $D$ happens, $B$ also happens. So $0.9$ of the red column will be dark ($B$), and $0.1$ will be light:

>However, 30% of the patients without Diseasitis will also turn the tongue depressor black.

So we break up the blue $\neg D$ column by $\bP(B \mid \neg D) = 0.3$ and $\bP(\neg B \mid \neg D) = 0.7$:

Now we would like to know the probability $\bP(D \mid B)$ of Diseasitis once we've observed that the tongue depressor is black. Let's break up our diagram by whether or not $B$ happens:

Conditioning on $B$ is like only looking at the part of our distribution where $B$ happens. So the probability $\bP(D \mid B)$ of $D$  conditioned on $B$ is the proportion of that area where $D$ also happens:

Here we can see why $\bP(D \mid B)$ isn't all that big. It's true that $\bP(B,D)$ is big relative to $\bP(\neg B,D)$, since we know that $\bP(B \mid D)$ is big (patients with Diseasitis almost always have black tongue depressors):

But this ratio doesn't really matter if we want to know $\bP(D \mid B)$, the probability that a patient with a black tongue depressor has Diseasitis. What matters is that we also assign a reasonably high probability $\bP(B, \neg D)$ to the patient having a black tongue depressor but nevertheless *not* suffering from Diseasitis:

So even when we see a black tongue depressor, there's still a pretty high chance the patient is healthy anyway, and our [https://arbital.com/p/-1rp](https://arbital.com/p/-1rp) $\bP(D\mid B)$ is not that high. Recall our square of probabilities:

When asked about $\bP(D\mid B)$, we think of the really high probability $\bP(B\mid D) = 0.9$:

Really, we should look at the part of our probability mass where $B$ happens, and see that a sizeable portion goes to places where $\neg D$ happens, and the patient is healthy:

Side note
---

The square visualization is very similar to [frequency diagrams](https://arbital.com/p/1x1), except we can just think in terms of probability mass rather than specifically frequency. Also, see that page for [waterfall diagrams](https://arbital.com/p/1x1), another way to visualize updating probabilities.


Computer Programming Familiarity

This page exists so that math explanations can be customized on the basis of whether or not the readers is familiar with coding. If you know how to code, then click the three dot icon and check the "this requisite" box.


Group conjugate

Two elements $x, y$ of a [https://arbital.com/p/-3gd](https://arbital.com/p/-3gd) $G$ are *conjugate* if there is some $h \in G$ such that $hxh^{-1} = y$.

# Conjugacy as "changing the worldview"

Conjugating by $h$ is equivalent to "viewing the world through $h$'s eyes".
This is most easily demonstrated in the [https://arbital.com/p/-497](https://arbital.com/p/-497), where it [is a fact](https://arbital.com/p/4bh) that if $$\sigma = (a_{11} a_{12} \dots a_{1 n_1})(a_{21} \dots a_{2 n_2}) \dots (a_{k 1} a_{k 2} \dots a_{k n_k})$$
and $\tau \in S_n$, then $$\tau \sigma \tau^{-1} = (\tau(a_{11}) \tau(a_{12}) \dots \tau(a_{1 n_1}))(\tau(a_{21}) \dots \tau(a_{2 n_2})) \dots (\tau(a_{k 1}) \tau(a_{k 2}) \dots \tau(a_{k n_k}))$$

That is, conjugating by $\tau$ has "caused us to view $\sigma$ from the point of view of $\tau$".

Similarly, in the [https://arbital.com/p/-4cy](https://arbital.com/p/-4cy) $D_{2n}$ on $n$ vertices, conjugation of the rotation by a reflection yields the inverse of the rotation: it is "the rotation, but viewed as acting on the reflected polygon".
Equivalently, if the polygon is sitting on a glass table, conjugating the rotation by a reflection makes the rotation act "as if we had moved our head under the table to look upwards first".

In general, if $G$ is a group which [acts](https://arbital.com/p/3t9) as (some of) the symmetries of a certain object $X$ %%note:Which [we can always view as being the case](https://arbital.com/p/49b).%% then conjugation of $g \in G$ by $h \in G$ produces a symmetry $hgh^{-1}$ which acts in the same way as $g$ does, but on a copy of $X$ which has already been permuted by $h$.

# Closure under conjugation

If a subgroup $H$ of $G$ is closed under conjugation by elements of $G$, then $H$ is a [https://arbital.com/p/-4h6](https://arbital.com/p/-4h6).
The concept of a normal subgroup is extremely important in group theory.

%%%knows-requisite([https://arbital.com/p/3t9](https://arbital.com/p/3t9)):
# Conjugation action

Conjugation forms a [action](https://arbital.com/p/3t9).
Formally, let $G$ act on itself: $\rho: G \times G \to G$, with $\rho(g, k) = g k g^{-1}$.
It is an exercise to show that this is indeed an action.
%%hidden(Show solution):
We need to show that the identity acts trivially, and that products may be broken up to act individually.

- $\rho(gh, k) = (gh)k(gh)^{-1} = ghkh^{-1}g^{-1} = g \rho(h, k) g^{-1} = \rho(g, \rho(h, k))$;
- $\rho(e, k) = eke^{-1} = k$.
%%

The [stabiliser](https://arbital.com/p/group_stabiliser) of this action, $\mathrm{Stab}_G(g)$ for some fixed $g \in G$, is the set of all elements such that $kgk^{-1} = g$: that is, such that $kg = gk$.
Equivalently, it is the [centraliser](https://arbital.com/p/group_centraliser) of $g$ in $G$: it is the subgroup of all elements which commute with $G$.

The [orbit](https://arbital.com/p/group_orbit) of the action, $\mathrm{Orb}_G(g)$ for some fixed $g \in G$, is the [https://arbital.com/p/-4bj](https://arbital.com/p/-4bj) of $g$ in $G$.
By the [https://arbital.com/p/-4l8](https://arbital.com/p/-4l8), this immediately gives that the size of a conjugacy class divides the [order](https://arbital.com/p/3gg) of the parent group.
%%%


There is only one logarithm

When you learned that we use [https://arbital.com/p/-decimal_notation](https://arbital.com/p/-decimal_notation) to record numbers, you may have wondered: If we used a different number system, such as a number system with 12 symbols instead of 10, how would that change the cost of writing numbers down? Are some number systems more efficient than others? On this page, we'll answer that question, among others.

In [https://arbital.com/p/4bz](https://arbital.com/p/4bz), we saw that any $f$ with [domain](https://arbital.com/p/3js) [$\mathbb R^+$](https://arbital.com/p/positive_reals) that satisfies the equation $f(x \cdot y) = f(x) + f(y)$ for all $x$ and $y$ in its domain is either trivial (i.e., it sends all inputs to 0), or it is [isomorphic](https://arbital.com/p/4f4) to [$\log_b$](https://arbital.com/p/3nd) for some base $b$. Thus, if we want a function's output to change by a constant that depends on $y$ every time its input changes by a factor of $y$, we only have one meaningful degree of freedom, and that's the choice of $b \neq 1$ such that $f(b) = 1$. Once we choose which value of $b$ $f$ sends to 1, the entire behavior of the function is fully defined.

How much freedom does this choice give us? Almost none! To see this, let's consider the difference between choosing base $b$ as opposed to an alternative base $c.$ Say we have an input $x \in \mathbb R^+$ — what's the difference between $\log_b(x)$ and $\log_c(x)?$

Well, $x = c^y$ for some $y$, because $x$ is positive. Thus, $\log_c(x)$ $=$ $\log_c(c^y)$ $=$ $y.$ By contrast, $\log_b(x)$ $=$ $\log_b(c^y)$ $=$ $y \log_b(c).$ ([Refresher](https://arbital.com/p/4bz).) Thus, $\log_c$ and $\log_b$ disagree on $x$ only by a constant factor — namely, $\log_b(c).$ And this is true for _any_ $x$ — you can get $\log_c(x)$ by calculating $\log_b(x)$ and dividing by $\log_b(c):$ $$\log_c(x) = \frac{\log_b(x)}{\log_b(c)}.$$

This is a remarkable equation, in the sense that it's worth remarking on. No matter what base $b$ we choose for $f$, and no matter what input $x$ we put into $f$, if we want to figure out what we would have gotten if we chose base $c$ instead, all we need to do is calculate $f(c)$ and divide $f(x)$ by $f(c).$ In other words, the different logarithm functions actually aren't very different at all — each one has all the same information as all the others, and you can recover the behavior of $\log_c$ using $\log_b$ and a simple calculation!

(By a symmetric argument, you can show that $\log_b(x) = \frac{\log_c(x)}{\log_c(b)},$ which implies that $\log_b(c) = \frac{1}{\log_c(b)}$, a fact we already knew from from studying [https://arbital.com/p/-44l](https://arbital.com/p/-44l).)

Above, we asked:

> When you learned that we use [https://arbital.com/p/-decimal_notation](https://arbital.com/p/-decimal_notation) to record numbers, you may have wondered: If we used a different number system, such as a number system with 12 symbols instead of 10, how would that change the cost of writing numbers down? Are some number systems more efficient than others?

From the fact that the [length of a written number grows logarithmically with the magnitude of the number](https://arbital.com/p/416) and the above equation, we can see that, no matter how large a number is, its base 10 representation differs in length from its base 12 representation only by a factor of $\log_{10}(12) \approx 1.08$. Similarly, the [binary](https://arbital.com/p/binary_notation) representation of a number is always about $\log_2(10) \approx 3.32$ times longer than its decimal representation. Because there is only one logarithm function (up to a multiplicative constant), which number base you use to represent numbers only affects the size of the representation by a multiplicative constant.

Similarly, if you ever want to convert between logarithmic measurements in different bases, you only need to perform a single multiplication (or division). For example, if someone calculated how many hours it took for the bacteria colony to triple, and you want to know how long it took to double, all you need to do is multiply by $\log_3(2) \approx 0.63.$ There is essentially only one logarithm function; the base merely defines the unit of measure. Given measurements taken in one base, you can easily convert to another.

Which base of the logarithm is best to use? Well, by the arguments above, it doesn't really matter which base you pick — it's easy enough to convert between bases if you need to. However, in computer science, [https://arbital.com/p/-3qq](https://arbital.com/p/-3qq), and many parts of physics, it's common to choose base 2. Why? Two reasons. First, because if you're designing a physical system to store data, a system that only has two stable states is the simplest available. Second, because if you need to pick an arbitrary amount that a thing has grown to call "one growth," then doubling (i.e., "it's now once again as large as it was when it started") is a pretty natural choice. (For example, many people find the 'half life' of a radioactive element to be a more intuitive notion than its 'fifth life,' even though both are equally valid measures of radioactivity.)

In other parts of physics and engineering, the log base 10 is more common, because it has a natural relationship to the way we represent numbers (using a base 10 representation). That is, $\log_{10}$ is one of the easiest logarithms to calculate in your head (because you can just [count the number of digits in a number's representation](https://arbital.com/p/416)), so it's often used by engineers.

Among mathematicians, the most common base of the logarithm is [$e$](https://arbital.com/p/e) $\approx 2.718,$ a [https://arbital.com/p/-trancendental_number](https://arbital.com/p/-trancendental_number) that cannot be fully written down. Why this strange base? That's an [advanced topic](https://arbital.com/p/base_e_is_natural) not covered in this tutorial. In brief, though, if you look at the [derivative](https://arbital.com/p/47d) of the logarithm (i.e., the way the function changes if you perturb the input), if the input was $x$ and you perturb it a tiny amount, the output changes by about a factor of $\frac{1}{x}$ times a constant $y$ that depends on the base of the logarithm, and $e$ is the base that causes that constant to be $1.$ For this reason and a handful of others, there is a compelling sense in which $\log_e$ is the "true form" of the logarithm. (So much so that it has its own symbol, $\ln,$ which is called the "natural logarithm.")

Regardless of which base you choose, converting between bases is easy — and as equation (1) shows, you only need to have one logarithm function in your toolkit in order to understand any other logarithm. Because if you have $\log_b$ and you want to know $\log_c(x)$, you can just calculate $\log_b(x)$ and divide it by $\log_b(c)$, and then you're done. In other words, there is really only one logarithm function, and it doesn't matter which one you pick.

[A visualization of the graph of the logarithm where you can change the base, but it changes only the labels on the axes while the curve itself remains completely static.](https://arbital.com/p/fixme:)

Nate Soares
Eric Rogstad
Patrick Stevens


The log lattice

[https://arbital.com/p/45q](https://arbital.com/p/45q) and other pages give physical interpretations of what logarithms are really doing. Now it's time to understand the raw numerics. Logarithm functions often output irrational, [https://arbital.com/p/-transcendental](https://arbital.com/p/-transcendental) numbers. For example, $\log_2(3)$ starts with

1.5849625007211561814537389439478165087598144076924810604557526545410982277943585625222804749180882420909806624750591673437175524410609248221420839506216982994936575922385852344415825363027476853069780516875995544737266834624612364248850047581810676961316404807130823233281262445248670633898014837234235783662478390118977006466312634223363341821270106098049177472541357330110499026268818251703576994712157113638912494135752192998699040767081539505404488360

What do these long strings of digits mean? Why these numbers, in particular? We already have the tools to answer those questions, all that remains is to put them together into a visualization. In [https://arbital.com/p/44l](https://arbital.com/p/44l), we saw why it is that $\log_2(3)$ must be this number: The cost of a [3-digit](https://arbital.com/p/4sj) in terms of 2-digits is more than 1 and less than 2; and the cost of ten 3-digits = one $3^{10}$-digit in terms of 2-digits is more than 15 and less than 16, and the cost of a hundred 3-digits = one $3^{100}$-digit in terms of 2-digits is more than 158 and less than 159, and so on. The giant number above is telling us about the _entire sequence_ of $3^{n}$-digit costs, for every $n$. If we double it, we get the cost of a 9-digit in terms of 2-digits. If we triple it, we get the cost of a 27-digit in terms of 2-digits.

In fact, this number also interacts correctly with all the other outputs of $\log_2.$ $\log_2(2)=1,$ and if we add 1 to the number above, we get the cost of a 6-digit in terms of 2-digits.

That is, the outputs of the log base 2 form a gigantic lattice, with an often-irrational output corresponding to each input. The gigantic lattice is set up just so, such that whenever you start at $x$ on the left and go to $x \cdot y,$ the output you get on the right is the output to that corresponds to $x$ plus the output that corresponds to $y$.

![The values of log2 on integers from 1 to 10](http://i.imgur.com/FvRymOk.png)

_The values of the logarithm base 2 on integers from 1 to 10._

The log lattice is this giant lattice assigning a specific number to each number, such that multiplying on the left corresponds to adding on the right.

![A demonstration that log2(3^10) is 10 log2(3)](http://i.imgur.com/tuTYlge.png)

_An illustration of some of the connections between the logs base 2 of powers of 3._

It's an intricate lattice that preservers a huge amount of structure — all the structure of the numbers, in fact, given that $\log$ is invertible. $\log_2(3)$ is a number that is simultaneously $1$ less than $\log_2(6),$ and half of $\log_2(9)$, and a tenth of $\log_2(3^{10}),$ and $\log_2(3^9)$ less than $\log_2(3^{10}).$ The value of $\log_2(3)$ has to satisfy a _massive_ number of constraints, in order to be precisely the number such that multiplication on the left corresponds to addition on the right. It's no surprise, then, that [it has no concise decimal expansion](https://arbital.com/p/4n8).

![A demonstration of the relationships between the log2 powers of 3](http://i.imgur.com/e650J5j.png)

_Notice the connections between the logs base 2 of the powers of 3. log 2 of 3 is a tenth of log 2 of 310 is a tenth of the log 2 of 3100, which is twice the log 2 of 350 which is twice the log 2 of 325. And, of course, the log 2 of 3101 is about 1.58 units larger than the log 2 of 3100._

In fact, the constraints on this log lattice are so tight that there's only one way to do it, up to a multiplicative constant. You can multiply everything on the right side by a constant, and that's the only thing you can do without disturbing the structure. For example, here's part of the log lattice viewed in base 3:

![The base 3 log lattice](http://i.imgur.com/yhcy9BZ.png)

It's exactly the same as the log lattice viewed in base 2, except that everything on the right has been divided by about 1.58 (i.e., multiplied by about 0.63).

What are logarithms doing? Well, fundamentally, there is a way to transform numbers such that what was once multiplication is now addition. If you were going to do multiplication to some numbers, you can transform them into this lattice, and then do addition, and then transform them back, and you'll get the same result. There's only one way to preserve all that structure, although if you want to view the structure, you need to (arbitrarily) choose some number on the left to call "1" on the right. Given that choice $b$ of translation, the log base $b$ taps into that intricate lattice.

[https://arbital.com/p/visualization](https://arbital.com/p/visualization)

Malcolm McCrimmon
Nate Soares
Alexei Andreev
Eric Rogstad


Formal definition

Meta tag for pages which give formal or brief jargon-heavy technical definitions of a concept. Formal definitions should be secondary [lenses](https://arbital.com/p/17b) when explanations are available.

Eric Bruylant
Alexei Andreev
Jessica Taylor
Patrick Stevens


Life in logspace

[https://arbital.com/p/4gp](https://arbital.com/p/4gp) hints at the reason that engineers, scientists, and AI researchers find logarithms so useful.

At this point, you may have a pretty good intuition for what logarithms are doing, and you may be saying to yourself

> Ok, I understand logarithms, and I see how they could be a pretty useful concept sometimes. Especially if, like, someone happens to have an out-of-control bacteria colony lying around. I can also see how they're useful for measuring representations of things (like the length of a number when you write it down, or the cost of communicating a message). If I squint, I can _sort of_ see why it is that the brain encodes many percepts logarithmically (such that the tone that I perceive is the $\log_2$ of the frequency of vibration in the air), as it's not too surprising that evolution occasionally hit upon this natural tool for representing things. I guess logs are pretty neat.

Then you start seeing logarithms in other strange places — say, you learn that [AlphaGo](https://en.wikipedia.org/wiki/AlphaGo) primarily manipulated [logarithms of the probability](https://arbital.com/p/log_probability) that the human would take a particular action instead of manipulating the probabilities directly. Then maybe you learn that scientists, engineers, and navigators of yore used to pay large sums of money for giant, pre-computed tables of logarithm outputs.

At that point, you might say

> Hey, wait, why do these logs keep showing up all over the place? Why are they _that_ ubiquitous? What's going on here?

The reason here is that _addition is easier than multiplication,_ and this is true both for navigators of yore (who had to do lots of calculations to keep their ships on course), and for modern AI algorithms.

Let's say you have a whole bunch of numbers, and you know you're going to have to perform lots and lots of operations on them that involve taking some numbers you already have and multiplying them together. There are two ways you could do this: One is that you could bite the bullet and multiply them all, which gets pretty tiresome (if you're doing it by hand) and/or pretty expensive (if you're training a gigantic neural network on millions of games of go). Alternatively, you could put all your numbers through a log function (a one-time cost), and then perform all your operations using addition (much cheaper!), and then transform them back using [https://arbital.com/p/-exponentiation](https://arbital.com/p/-exponentiation) when you're done (another one-time cost).

Empirically, the second method tends to be faster, cheaper, and more convenient (at the cost of some precision, given that most log outputs are [transcendental](https://arbital.com/p/transcendental_number)).

This is the last insight about logarithms contained in this tutorial, and it is a piece of practical advice: If you have to perform a large chain of multiplications, you can often do it cheaper by first transforming your problem into "log space", where multiplication acts like addition (and exponentiation acts like multiplication). Then you can perform a bunch of additions instead, and transform the answer back into normal space at the end.

Nate Soares
Eric Rogstad
Alexei Andreev
Malcolm McCrimmon


The End (of the basic log tutorial)

That concludes our introductory tutorial on logarithms! You have made it to the end.

Throughout this tutorial, we saw that the logarithm base $b$ of $x$ calculates the number of $b$-factors in $x.$ Hopefully, this claim now means more to you than it once did. We've seen a number of different ways of interpreting what logarithms are doing, including:

- $\log_b(x) = y$ means ["it takes about $y$ digits to write $x$ in base $b$."](https://arbital.com/p/416)
- $\log_b(x) = y$ means ["it takes about $y$ $b$-digits to emulate an $x$-digit."](https://arbital.com/p/44l)
- $\log_b(x) = y$ means ["if the space of possible messages to send goes up by a factor of $x$, then the cost, in $b$-digits, goes up by a factor of $y$](https://arbital.com/p/45q)
- And, simply, $\log_b(x) = y$ means that if you start with 1 and grow it by factors of $b$, then after $y$ iterations of this your result will be $x.$

For example, $\log_2(100)$ counts the number of doublings that constitute a factor-of-100 increase. (The answer is more than 6 doublings, but slightly less than 7 doublings).

We've also seen that any function $f$ whose output grows by a constant (that depends on $y$) every time its input grows by a factor of $y$ is [very likely a logarithm function](https://arbital.com/p/4bz), and that, in essence, [https://arbital.com/p/-4gm](https://arbital.com/p/-4gm) function.

We've glanced at the [underlying structure](https://arbital.com/p/4gp) that all logarithm functions tap into, and we've briefly discussed [what makes working with logarithms so dang useful](https://arbital.com/p/4h0).

There are also a huge number of questions about, applications for, and extensions of the logarithm that we _didn't_ explore. Those include, but are not limited to:

- Why is $e$ the natural base of the logarithm?
- What is up with the link between logarithms, exponentials, and roots?
- What is the derivative of $\log_b(x)$ and why is it proportional to $\frac{1}{x}$?
- How can logarithms be efficiently calculated?
- What happens when we extend logarithms to complex numbers, and why is the result a [https://arbital.com/p/-multifunction](https://arbital.com/p/-multifunction)?

Answering these questions will require an advanced tutorial on logarithms. Such a thing does not exist yet, but you can help make it happen.


Normal subgroup

A *normal subgroup* $N$ of group $G$ is one which is closed under [conjugation](https://arbital.com/p/4gk): for all $h \in G$, it is the case that $\{ h n h^{-1} : n \in N \} = N$.
In shorter form, $hNh^{-1} = N$.

Since [conjugacy is equivalent to "changing the worldview"](https://arbital.com/p/4gk), a normal subgroup is one which "looks the same from the point of view of every element of $G$".

A subgroup of $G$ is normal if and only if it is the [kernel](https://arbital.com/p/49y) of some [https://arbital.com/p/-47t](https://arbital.com/p/-47t) from $G$ to some group $H$. ([Proof.](https://arbital.com/p/4h7))

%%%knows-requisite([https://arbital.com/p/category_theory_equaliser](https://arbital.com/p/category_theory_equaliser)):
From a category-theoretic point of view, the kernel of $f$ is an equaliser of an arrow $f$ with the zero arrow; it is therefore universal such that composition with $f$ yields zero.
%%%


Subgroup is normal if and only if it is the kernel of a homomorphism

Let $N$ be a [https://arbital.com/p/-subgroup](https://arbital.com/p/-subgroup) of [https://arbital.com/p/-3gd](https://arbital.com/p/-3gd) $G$.
Then $N$ is [normal](https://arbital.com/p/4h6) in $G$ if and only if there is a group $H$ and a [https://arbital.com/p/-47t](https://arbital.com/p/-47t) $\phi:G \to H$ such that the [kernel](https://arbital.com/p/49y) of $\phi$ is $N$.

# Proof

## "Normal" implies "is a kernel"
Let $N$ be normal, so it is closed under [conjugation](https://arbital.com/p/4gk).
Then we may define the [https://arbital.com/p/-quotient_group](https://arbital.com/p/-quotient_group) $G/N$, whose elements are the [left cosets](https://arbital.com/p/group_coset) of $N$ in $G$, and where the operation is that $gN + hN = (g+h)N$.
This group is well-defined ([proof](https://arbital.com/p/quotient_by_subgroup_is_well_defined_iff_normal)).

Now there is a homomorphism $\phi: G \to G/N$ given by $g \mapsto gN$.
This is indeed a homomorphism, by definition of the group operation $gN + hN = (g+h)N$.

The kernel of this homomorphism is precisely $\{ g : gN = N \}$; that is simply $N$:

- Certainly $N \subseteq \{ g : gN = N \}$ (because $nN = N$ for all $n$, since $N$ is closed as a subgroup of $G$);
- We have $\{ g : gN = N \} \subseteq N$ because if $gN = N$ then in particular $g e \in N$ (where $e$ is the group identity) so $g \in N$.

## "Is a kernel" implies "normal"
Let $\phi: G \to H$ have kernel $N$, so $\phi(n) = e$ if and only if $n \in N$.
We claim that $N$ is closed under conjugation by members of $G$.

Indeed, $\phi(h n h^{-1}) = \phi(h) \phi(n) \phi(h^{-1}) = \phi(h) \phi(h^{-1})$ since $\phi(n) = e$.
But that is $\phi(h h^{-1}) = \phi(e)$, so $hnh^{-1} \in N$.

That is, if $n \in N$ then $hnh^{-1} \in N$, so $N$ is normal.


Quotient by subgroup is well defined if and only if subgroup is normal

Let $G$ be a [https://arbital.com/p/-3gd](https://arbital.com/p/-3gd) and $N$ a [https://arbital.com/p/-4h6](https://arbital.com/p/-4h6) of $G$.
Then we may define the *quotient group* $G/N$ to be the set of [left cosets](https://arbital.com/p/4j4) $gN$ of $N$ in $G$, with the group operation that $gN + hN = (gh)N$.
This is well-defined if and only if $N$ is normal.

# Proof

## $N$ normal implies $G/N$ well-defined

Recall that $G/N$ is well-defined if "it doesn't matter which way we represent a coset": whichever coset representatives we use, we get the same answer.

Suppose $N$ is a normal subgroup of $G$.
We need to show that given two representatives $g_1 N = g_2 N$ of a coset, and given representatives $h_1 N = h_2 N$ of another coset, that $(g_1 h_1) N = (g_2 h_2)N$.

So given an element of $g_1 h_1 N$, we need to show it is in $g_2 h_2 N$, and vice versa.

Let $g_1 h_1 n \in g_1 h_1 N$; we need to show that $h_2^{-1} g_2^{-1} g_1 h_1 n \in N$, or equivalently that $h_2^{-1} g_2^{-1} g_1 h_1 \in N$.

But $g_2^{-1} g_1 \in N$ because $g_1 N = g_2 N$; let $g_2^{-1} g_1 = m$.
Similarly $h_2^{-1} h_1 \in N$ because $h_1 N = h_2 N$; let $h_2^{-1} h_1 = p$.

Then we need to show that $h_2^{-1} m h_1 \in N$, or equivalently that $p h_1^{-1} m h_1 \in N$.

Since $N$ is closed under conjugation and $m \in N$, we must have that $h_1^{-1} m h_1 \in N$;
and since $p \in N$ and $N$ is closed under multiplication, we must have $p h_1^{-1} m h_1 \in N$ as required.

## $G/N$ well-defined implies $N$ normal

Fix $h \in G$, and consider $hnh^{-1} N + hN$.
Since the quotient is well-defined, this is $(hnh^{-1}h) N$, which is $hnN$ or $hN$ (since $nN = N$, because $N$ is a subgroup of $G$ and hence is closed under the group operation).
But that means $hnh^{-1}N$ is the identity element of the quotient group, since when we added it to $hN$ we obtained $hN$ itself.

That is, $hnh^{-1}N = N$.
Therefore $hnh^{-1} \in N$.

Since this reasoning works for any $h \in G$, it follows that $N$ is closed under conjugation by elements of $G$, and hence is normal.


Alternating group

The *alternating group* $A_n$ is defined as a certain subgroup of the [https://arbital.com/p/-497](https://arbital.com/p/-497) $S_n$: namely, the collection of all elements which can be made by multiplying together an even number of [transpositions](https://arbital.com/p/4cn).
This is a well-defined notion ([proof](https://arbital.com/p/4hg)).

%%%knows-requisite([https://arbital.com/p/4h6](https://arbital.com/p/4h6)):
$A_n$ is a [https://arbital.com/p/-4h6](https://arbital.com/p/-4h6) of $S_n$; it is the [quotient](https://arbital.com/p/quotient_group) of $S_n$ by the [sign homomorphism](https://arbital.com/p/4hk).
%%%

# Examples

- A [cycle](https://arbital.com/p/49f) of even length is an odd permutation in the sense that it can only be made by multiplying an odd number of transpositions.
For example, $(132)$ is equal to $(13)(23)$.
- A cycle of odd length is an even permutation, in that it can only be made by multiplying an even number of transpositions.
For example, $(1354)$ is equal to $(54)(34)(14)$.

- The alternating group $A_4$ consists precisely of twelve elements: the identity, $(12)(34)$, $(13)(24)$, $(14)(23)$, $(123)$, $(124)$, $(134)$, $(234)$, $(132)$, $(143)$, $(142)$, $(243)$.

# Properties

%%%knows-requisite([https://arbital.com/p/4h6](https://arbital.com/p/4h6)):
The alternating group $A_n$ is of [index](https://arbital.com/p/index_of_a_subgroup) $2$ in $S_n$.
Therefore $A_n$ is [normal](https://arbital.com/p/4h6) in $S_n$ ([proof](https://arbital.com/p/4hl)).
Alternatively we may give the homomorphism explicitly of which $A_n$ is the [kernel](https://arbital.com/p/49y): it is the [sign homomorphism](https://arbital.com/p/4hk).
%%%

- $A_n$ is generated by its $3$-cycles. ([Proof.](https://arbital.com/p/4hs))
- $A_n$ is [simple](https://arbital.com/p/4jc). ([Proof.](https://arbital.com/p/4jb))
- The [conjugacy classes](https://arbital.com/p/4bj) of $A_n$ are [easily characterised](https://arbital.com/p/alternating_group_five_conjugacy_classes).


The collection of even-signed permutations is a group

The collection of elements of the [https://arbital.com/p/-497](https://arbital.com/p/-497) $S_n$ which are made by multiplying together an even number of permutations forms a subgroup of $S_n$.

This proves that the [https://arbital.com/p/-alternating_group](https://arbital.com/p/-alternating_group) $A_n$ is well-defined, if it is given as "the subgroup of $S_n$ containing precisely that which is made by multiplying together an even number of transpositions".

# Proof

Firstly we must check that "I can only be made by multiplying together an even number of transpositions" is a well-defined notion; this [is in fact true](https://arbital.com/p/4hh).

We must check the group axioms.

- Identity: the identity is simply the product of no transpositions, and $0$ is even.
- Associativity is inherited from $S_n$.
- Closure: if we multiply together an even number of transpositions, and then a further even number of transpositions, we obtain an even number of transpositions.
- Inverses: if $\sigma$ is made of an even number of transpositions, say $\tau_1 \tau_2 \dots \tau_m$, then its inverse is $\tau_m \tau_{m-1} \dots \tau_1$, since a transposition is its own inverse.


The sign of a permutation is well-defined

The [https://arbital.com/p/-497](https://arbital.com/p/-497) $S_n$ contains elements which are made up from [transpositions](https://arbital.com/p/4cn) ([proof](https://arbital.com/p/4cp)).
It is a fact that if $\sigma \in S_n$ can be made by multiplying together an even number of transpositions, then it cannot be made by multiplying an odd number of transpositions, and vice versa.

%%%knows-requisite([https://arbital.com/p/47y](https://arbital.com/p/47y)):
Equivalently, there is a [https://arbital.com/p/-47t](https://arbital.com/p/-47t) from $S_n$ to the [https://arbital.com/p/-47y](https://arbital.com/p/-47y) $C_2 = \{0,1\}$, taking the value $0$ on those permutations which are made from an even number of permutations and $1$ on those which are made from an odd number.
%%%

# Proof

Let $c(\sigma)$ be the number of cycles in the [disjoint cycle decomposition](https://arbital.com/p/49f) of $\sigma \in S_n$, including singletons.
For example, $c$ applied to the identity yields $n$, while $c((12)) = n-1$ because $(12)$ is shorthand for $(12)(3)(4)\dots(n-1)(n)$.
We claim that multiplying $\sigma$ by a transposition either increases $c(\sigma)$ by $1$, or decreases it by $1$.

Indeed, let $\tau = (kl)$.
Either $k, l$ lie in the same cycle in $\sigma$, or they lie in different ones.

- If they lie in the same cycle, then $$\sigma = \alpha (k a_1 a_2 \dots a_r l a_s \dots a_t) \beta$$ where $\alpha, \beta$ are disjoint from the central cycle (and [hence commute](https://arbital.com/p/49g) with $(kl)$).
Then $\sigma (kl) = \alpha (k a_s \dots a_t)(l a_1 \dots a_r) \beta$, so we have broken one cycle into two.
- If they lie in different cycles, then $$\sigma = \alpha (k a_1 a_2 \dots a_r)(l b_1 \dots b_s) \beta$$ where again $\alpha, \beta$ are disjoint from $(kl)$.
Then $\sigma (kl) = \alpha (k b_1 b_2 \dots b_s l a_1 \dots a_r) \beta$, so we have joined two cycles into one.

Therefore $c$ takes even values if there are evenly many transpositions in $\sigma$, and odd values if there are odd-many transpositions in $\sigma$.

More formally, let $\sigma = \alpha_1 \dots \alpha_a = \beta_1 \dots \beta_b$, where $\alpha_i, \beta_j$ are transpositions.
%%%knows-requisite([https://arbital.com/p/modular_arithmetic](https://arbital.com/p/modular_arithmetic)):
(The following paragraph is more succinctly expressed as: "$c(\sigma) \equiv n+a \pmod{2}$ and also $\equiv n+b \pmod{2}$, so $a \equiv b \pmod{2}$.")
%%%
Then $c(\sigma)$ flips odd-to-even or even-to-odd for each integer $1, 2, \dots, a$; it also flips odd-to-even or even-to-odd for each integer $1, 2, \dots, b$.
Therefore $a$ and $b$ must be of the same [parity](https://arbital.com/p/even_odd_parity).


Isomorphism: Intro (Math 0)

If two things are essentially the same from a certain perspective, and they only differ in unimportant details, then they are **isomorphic**.

##Comparing Amounts##
Consider the Count von Count. He cares only about counting things. He doesn't care what they are, just how many there are. He decides that he wants to collect items into plastic crates, and he considers two crates equal if both contain the same number of items.

![Equivalent Crates](http://i.imgur.com/sG7Qfyv.jpg)

Now Elmo comes to visit, and he wants to impress the Count, but Elmo is not great at counting. Without counting them explicitly, how can Elmo tell if two crates contain the same number of items?

Well, he can take one item out of each crate and put the pair to one side.

![Pairing one pair of items](https://i.imgur.com/myTqIrRl.jpg)

 He continues pairing items up in this way and when one crate runs out he checks if there are any left over in the other crate. If there aren't any left over, then he knows there were the same number of items in both crates.

![Pairing all items in one crate with items in the other](https://i.imgur.com/S87JO2nl.jpg)

Since the Count von Count only cares about counting things, the two crates are basically equivalent, and might as well be the same crate to him. Whenever two objects are the same from a certain perspective, we say that they are **isomorphic**.

In this example, the way in which the crates were the same is that each item in one crate could be paired with an item in the other.

![Pairing all items in one box with items in the other](https://i.imgur.com/S87JO2nl.jpg)

![Bijection between crates](https://i.imgur.com/53YbraFl.jpg)

This wouldn't have been possible if the crates had different numbers of items in them.

![Different numbers of items](https://i.imgur.com/F5rLwAsl.jpg)

![No way to pair items](https://i.imgur.com/KCI9UOvl.jpg)

Whenever you can match each item in one collection with exactly one item in another collection, we say that the collections are **[bijective](https://arbital.com/p/-499)** and the way you paired them is a **[bijection](https://arbital.com/p/-499)**. A bijection is a specific kind of isomorphism.

![Bijection between crates](https://i.imgur.com/53YbraFl.jpg)

Note that there might be many different bijections between two bijective things.

![Another bijection between crates](http://i.imgur.com/Q6Si1ZX.png)

In fact, all that counting involves is pairing up the things you want to count, either with your fingers or with the concepts of 'numbers' in your head. If there are as many objects in one crate as there are numbers from one to seven, and there are as many objects in another crate as numbers from one to seven, then both crates contain the same number of objects.

##Comparing Maps##
Now imagine that you have a map of the London Underground. Such a map is not to scale, nor does it even show how the  tracks bend or which station is in which direction compared to another. They only record which stations are connected.

![Map of the London Underground](http://i.imgur.com/9d8oX2K.gif)

Your Spanish friend is coming to visit and you want to get them a version of the map in Spanish. But on the Spanish maps, the shape of the tracks is different and you can't read Spanish. What's more, not all the maps are of the London Underground! What do you do? Well, given a Spanish map and your English map, you can try to match up the stations (through trial and error) and if the stations are all *connected to each other in the same ways* on both maps then you know they are both of the same train system.

More precisely, consider the following smaller (fictional) example. There are stations *Trafalgal*, *Marybone*, *Oxbridge*, *Eastweston* and *Charlesburrough* on the English map. *Marybone* is connected to all the other stations, and *Charlesburrough* is connected to *Eastweston*. (There are no other connections)

![Fictional London Underground Map](http://i.imgur.com/J3p094x.png)

%note: We don't want to worry about whether stations are connected to themselves. You can just assume no station is ever connected to itself.%

%note: If one station is connected to another, then the second station is also connected to the first. So since *Marybone* is connected to *Eastweston* then *Eastweston* is connected to *Marybone*. If it seems silly to even mention this fact, then don't worry to much. It's just that we might just have easily decided that there's a one-way train running from *Marybone* to *Eastweston* but not in the other direction.%

Assume the Spanish map has the following stations: *Patata*, *Huesto*, *Carbon*, *Esteoeste*, and *Puente de Buey*. Assume also that *Huesto* is connected to every other station, and that *Carbon* is connected to *Esteoeste*. (Again, there are no other connections).

![Fictional Spanish Map](http://i.imgur.com/nENfDqk.png)

Then the two maps are essentially the same for your purposes. They are **isomorphic** (as [graphs](https://arbital.com/p/-graph), in fact), and the way that you matched the stations on the one map with those on the other is an **isomorphism**.

## Isomorphisms##
Imagine that you have a *London Underground Official Spanish-to-English Train Station Dictionary* that tells you how to translate the names of the train stations. So, for example, you can use it to convert *Patata* to *Trafalgal*. Then this dictionary is an **isomorphism** from the Spanish map to the English one.

In particular, the dictionary translates *Huesto* to *Marybone*, *Patata* to *Tralfalgal*, *Puento de Buey* to *Oxbridge*, *Esteoeste* to *Eastweston*, and *Carbon* to *Charlesburrough*.

You could also get a *London Underground Official* **English-to-Spanish** *Train Station Dictionary*. Then, if you were to use this dictionary to translate *Tralfalgal*, you'd get back *Patata*. Hence your first original translation of that station from English to Spanish has been undone.

In particular, the dictionary translates *Marybone* to *Huesto*, *Tralfalgal* to *Patata*. *Oxbridge* to *Puento de Buey*, *Eastweston* to *Esteoeste*, and *Charlesburrough* to *Carbon*.

![English Stations Paired Up with Spanish Stations](http://i.imgur.com/sElv72C.png)

In fact, if you take the English map and translate all of the stations into Spanish using the one dictionary and then translate back, you'd get back to where you started. Similarly, if you translated the Spanish map from Spanish into English with the one dictionary and back to Spanish with the other you'd get the Spanish map back. Hence both of these dictionaries are complete [inverses](https://arbital.com/p/-inverse) of each other.

In fact, in [category theory](https://arbital.com/p/-4c7), this is exactly the definition of an isomorphism: if you have some translation ([https://arbital.com/p/-4d8](https://arbital.com/p/-4d8)) such that you can find a backwards translation (morphism in the opposite direction), and using the one translation after the other is for all intents and purposed the same as not having translated anything at all (i.e., no important information is lost in translation), then the original translation is an **isomorphism**. (In fact, both of them are isomorphisms).

What if you had a different pair of dictionaries.

In particular, what if the Spanish-to-English dictionary translates *Huesto* to *Marybone*, *Puento de Buey* to *Tralfalgal*, *Patata* to *Oxbridge*, *Carbon* to *Eastweston*, and *Esteoeste* to *Charlesburrough*?

Then if we translate from Spanish into English, and then translate back with the original English-to-Spanish dictionary, then *Patata* is first translated to *Oxbridge*, but then it is translated back into *Puente de Buey*. So it does not reverse the translation. Hence this is not an isomorphism.

However, what if the English-to-Spanish Dictionary translates *Marybone* to *Huesto*, *Oxbridge* to *Patata*. *Eastweston* to *Puento de Buey*, *Tralfalgal* to *Esteoeste*, and *Charlesburrough* to *Carbon*? Then the translations are reverses of each other. Hence this is another isomorphism. There may be many isomorphisms between two isomorphic maps.

![Another Way of Pairing English Stations with Spanish Stations](http://i.imgur.com/X6A6oNb.png)

##Non-Isomorphic Maps##
Imagine you had the English map from above.  As a reminder the stations were:
*Marybone*, *Eastweston*, *Charlesburrough*, *Tralfalgal*, and *Oxbridge*.

If, now, you find a Spanish map with only four stations on it, it can't possibly be isomorphic to your English map; there would be some station appearing on the English map which isn't named on the Spanish one.

![Spanish Map with Only Four Stations When English Map has Five](http://i.imgur.com/BkVZBWB.png)

Similarly, if the Spanish map has six stations, then they aren't isomorphic either since there is an extra station on the Spanish map not appearing on the English one.

![Spanish Map with Six Stations When English Map has Five](http://i.imgur.com/EYYmzWi.png)

What if there are five stations on the Spanish map. Is it then definitely isomorphic to the English one?

Recall that:
*Marybone* is connected to every other station, and *Eastweston* is connected to *Charlesburrough*.

But what if now instead on the Spanish map, *Huesto* is still connected to everything, but nothing else is connected to anything else.

![Huesto is Connected to Everything With No Other Connections. English Map Still Has the Same Connections](http://i.imgur.com/4mQD0hK.png)

Then the translation taking *Marybone* to *Huesto*, *Tralfalgal* to *Patata*. *Oxbridge* to *Puento de Buey*, *Eastweston* to *Esteoeste*, and *Charlesburrough* to *Carbon* is not an isomorphism. *Eastweston* is connected to *Charlesburrough* on the English map, but the corresponding stations on the Spanish map, *Esteoeste* and *Carbon*, are not connected to each other. Hence this translation is not an isomorphism, since under these translations, the maps represent different things.

But even though this way of pairing up the stations isn't an isomorphism, maybe there is another way of pairing them up which is? But no, even this is doomed to failure because the *number of connections* in both cases is different. For example, *Eastweston* is connected to two stations, but no station on the Spanish map is connected to two other stations. Hence there cannot be any translation that works. No isomorphism exists between the two maps.

If no isomorphism exists between two structures, then they are **non-isomorphic**.

Notice, in fact, that the two maps do not have the same total number of connections. There are five connections on the English map, but only four on the Spanish map. Hence they cannot be isomorphic.

What if the Spanish map has the following connections? *Patata* to everything except *Carbon*, and *Carbon* to *Huesto*and *Esteoeste*. Then there still cannot be an isomorphism, since, again, *Marybone* is connected to four other stations, but nothing on the Spanish map is connected to four stations. In this case, both maps have five connections. Hence even if both maps have the same total number of connections, they may still be non-isomorphic.

![Both Maps Have Five Connections but Are Non-Isomorphic](http://i.imgur.com/BKXLPqv.png)

##Comparing Weights##
Not all isomorphisms need be mappings between structures. Consider if you work at the post-office and must weigh packages. You do not care about the size and shape of the packages, only their weight. Then you consider two packages isomorphic if their weights are equal.

Imagine, then, that you have two packages, say one containing a book %note: *The Official London Underground History of Train Stations*% and the other is a plastic crate %note: "To the Count, with love"%.

You also have a half-broken pair of brass scales: they have a pair of pans on which items can be placed.

![Empty Balanced Scales](http://i.imgur.com/2wuFlTO.png)

However, they can only tip to the left or remain flat.

![Empty Scales Tilting Left](http://i.imgur.com/oz3m8vF.png)

If the item on the left is heavier than the one on the right, then the scales tilt left.

![Scales With Heavy Object on Left, Scales Tilt Left](http://i.imgur.com/iSmYESb.png)

Otherwise, if they are of equal weight or the item on the left is lighter than the one on the right, the scales remain level.

![Scales With Same Object on Both Sides, Scales Flat](http://i.imgur.com/NZe1uQC.png)

![Scales With Heavy Object on Right, Scales Flat](http://i.imgur.com/0BqznlG.png)

Place the book on the left pan of the scale, and the crate on the right. If the scales balance then either the book is lighter than the crate or it is the same weight as the crate. Now swap them. If they remain level, then either the crate is lighter than the book or it is the same weight as the book. Since the book cannot be lighter than the crate whilst the crate is simultaneously lighter than the book, they must be the same weight. Hence they are isomorphic.

This very act of balancing the scales is an isomorphism. It has an inverse: just swap the two packages around! Start with the book on the left pan and the crate on the right. Then place the crate on the left pan and the book on the right. The fact that the scales balance both times tells you (the obvious fact) that the book weighs the same as itself. Since you already know this,  doing this actually tells you as much about the book's weight compared to itself as doing nothing at all.

If this last part seems silly or confusing, don't worry too much about it. It's just to illustrate how the idea of an isomorphism is intricately tied with having an inverse.

##An Isomorphism Joke##
A man walks into a bar. He is surprised at how the patrons are acting.  One of them says a number, like “forty-two”, and the rest break into laughter. He asks the bartender what’s going on. The bartender explains that they all come here so often that they’ve memorized all of each other’s jokes, and instead of telling them explicitly, they just give each a number, say the number, and laugh appropriately. The man is intrigued, so he shouts “Two thousand!”. He is shocked to find everyone laughs uproariously, the loudest he's heard that evening. Perplexed, he turns to the bartender and says  “They laughed so much more at mine than at any of the others." "Well of course," the bartender answers matter-of-factly, "they've never heard that one before!”

Eric Bruylant
Team Arbital
Eric Rogstad
Mark Chimes