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Looking for an example of a group homomorphism to better grasp the concept. Something along the lines of defining a simple group such as $X = \{1,0\}$ and $G = (X, \circ)$, and $A = \{1,2,3\}$ is an arbitrary set. Then taking that (or preferrably a better example) and demonstrating explicitly how the group homomorphism works by plugging in values to the function.

Background

This excerpt on group action says:

Thus, if $G$ is a group and $X$ is a set, then an action of $G$ on $X$ may be formally defined as a group homomorphism $\varphi$ from $G$ to the symmetric group of $X$.

The symmetric group is defined as:

the group whose elements are all the bijections from the set to itself, and whose group operation is the composition of functions.

A bijection is defined as:

a function between the elements of two sets, where each element of one set is paired with exactly one element of the other set, and each element of the other set is paired with exactly one element of the first set... A bijection from the set X to the set Y has an inverse function from Y to X.

A group homomorphism is defined as:

a function $h : G → H$ such that for all $u$ and $v$ in $G$ it holds that $h(u*v)=h(u)\cdot h(v)$, where the group operation on the left hand side of the equation is that of $G$ and on the right hand side that of $H$.

So we have a homomorphism is $\varphi : G \to \dot{X}$ let's say, where $\dot{X}$ is the symmetric group of X.

Question

I am wondering exactly what this looks like through plugging in examples to the function $\varphi$. The way I am currently looking at it, it seems that these are all equivalent definitions:

\begin{align*} \varphi_a &: G \to \dot{X}\\ \varphi_b &: G \to X \times X\\ \varphi_c &: G \to X \to X\\ \varphi_d &: G \times X \to X \end{align*}

Plugging in values would look something like this (I am very confused by this point):

\begin{align*} \varphi(1) &\mapsto 1\\ \varphi(1) &\mapsto (1,2)\\ \varphi(1) &\mapsto (1,2,3)\\ \varphi(1,2) &\mapsto 1\\ \varphi(1,2) &\mapsto (1,2)\\ \varphi(1,2) &\mapsto (1,2,3)\\ \end{align*}

(Don't know how many args it takes, how currying is applied here if at all, or what the output is)

My main source of confusion is in the definition of bijection and the symetric group:

  1. Is a bijection an ordered pair $((a,b),(b,a))$, or just $(a,b)$, or is it a function $f : a \mapsto b$, or should I interpret a function as simply an ordered pair and nothing else. (Coming from a programming background).
  2. What does it look like applying the group operation of a symmetric group (composition of functions) to a bijective function? Specifically, what are some example inputs/outputs in 1 or 2 different notational variations so I can get a sense of exactly what is meant.
  3. Combining it all together, what are some example inputs/outputs of a group homomorphism, having now seen examples of bijections and symmetry groups.

I am having difficulty visualizing how bijection -> symmetry group -> group homomorphism maps together, using values from an arbitrary example group and set for demonstration purposes.

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  • $\begingroup$ I'm totally confused by your question. Everything below the word "Question" is incomprehensible to me. What exactly is going on with your four "equivalent definitions"? What are $\phi_a$, $\phi_b$, and so on, and in what way are they "equivalent"? $\endgroup$ – Jack M Apr 21 '18 at 10:44
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I think defining the group action in terms of the symmetric group might be a little confusing. I will give you another definition and, as a consequence, I will try to show you what's the relationship with the symmetric group.

A (left) group action of a group $G$ over a set $X$ is commonly defined as a function

$\varphi:G\times X\to X$

$(g,x)\mapsto \varphi(g,x)$

Usually, one writes $\varphi(g,x)$ as $g\cdot x$ for convinience. The function must satisfy:

  • $e\cdot x=x$ where $e$ is the neutral element of $G$,
  • $(hg)\cdot x=h\cdot (g\cdot x)$ for every $h,g\in G$ (the absence of dot implies group operation here).

It follows that for every $g\in G$, the map $x\mapsto g\cdot x$ is bijective (with inverse $x\mapsto g^{-1}\cdot x$). Here is where you have the symmetric group. Then, the action becomes a group homomorphism between $G$ and $\dot{X}$ because it is the map that sends $e\mapsto e\cdot x=x$ (which can be identified with $e\mapsto (x\mapsto e\cdot x=x)$, hence $e$ maps to the identity) and similarly, $hg$ maps to the composition of $g\mapsto g\cdot x$ and $h\mapsto h\cdot x$.

In your example, $X$ has only two elements, therefore there are only two possible bijections, $Id$ (identity) and $(0\ 1)=\{0\mapsto 1, 1\mapsto 0\}$. Hence we can compute every possibility:

$\varphi(Id, 0)=Id(0)=0; $ $\varphi(Id,1)=Id(1)=1$

$\varphi((0\ 1),0)=1;$ $\varphi((0\ 1),1)=0$.

For finite sets it is convinient to think about bijections as permutations. As you can see, fixing the first argument gives you $\varphi(Id,*)=Id$ and $\varphi((0\ 1),*)=(0,1)$.

Let's try with $A=\{1,2,3\}$. Here there are $6$ bijections (permutations):

  1. $Id$
  2. $(1\ 2)=\{1\mapsto 2, 2\mapsto 1,3\mapsto 3\}$
  3. $(1\ 3)=\{1\mapsto 3, 2\mapsto 2,3\mapsto 1\}$
  4. $(2\ 3)=\{1\mapsto 1, 2\mapsto 3,3\mapsto 2\}$
  5. $(1\ 2\ 3)=\{1\mapsto 2, 2\mapsto 3, 3\mapsto 1\}$
  6. $(1\ 3\ 2)=\{1\mapsto 3, 2\mapsto 2,2\mapsto 1\}$.

I will compute a few examples.

$\varphi((1\ 3),x)=\begin{cases} 3 & \text{if }x=1\\ 2 & \text{if }x=2\\ 1 & \text{if }x=3 \end{cases}$

because it sends $(1\ 3)$ to $(1\ 3)\cdot x$ (or $x\mapsto (1\ 3)\cdot x)$), which for each value of $x$ gives a different image. Now, for a bijection that moves every element:

$\varphi((1\ 2\ 3),1)=(1\ 2\ 3)\cdot 1=2$ because $(1\ 2\ 3)$ sends $1$ to $2$, and so on.

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  • $\begingroup$ Ok so after first read I gather there are two ways of writing a bijection: $(0,1)={0↦1,1↦0}$. $\endgroup$ – Lance Pollard Apr 21 '18 at 10:19
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    $\begingroup$ Just to be clear, $(a,b)$ is the usual notation for an ordered pair. $(a~b)$ is the usual notation for a transposition. Since ordered pairs are under discussion here also, $(a,b)$ is a truly awful notation for a transposition. $\endgroup$ – C Monsour Apr 21 '18 at 10:23
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    $\begingroup$ Clever notation, I like how $(1\ 2)=\{1↦2,2↦1,3↦3\}$ you can leave off the 3. $\endgroup$ – Lance Pollard Apr 21 '18 at 10:33
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    $\begingroup$ @LancePollard the reason is that there are several points of view that are equivalent because the images are the same at the end. The idea is to see how the group "moves" the elements of the set, and you can formalize that in several ways. The notation $\varphi(1)(a)$ is very similar, you could interpret it as taking $1$ and $a$ as arguments and writing $\varphi(1,a)$, and similarly with mine, you could redefine it in order to write $\varphi(1\ 2\ 3)(1)$, since the main point is to obtain how the group acts on the set, i.e, what happens to the set once the group has acted. $\endgroup$ – Javi Apr 21 '18 at 10:41
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    $\begingroup$ @LancePollar a bijection is an element of a group, the symmetric group. $\endgroup$ – Javi Apr 21 '18 at 10:45
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(1) A bijection is a function that is 1-1 and onto. You can interpret it as a set of ordered pairs in which each element of the domain appears once as the first element of a pair, and each element of the codomain appears once as the second element of a pair.

(2) The symmetric group of $X$ is the collection of bijections from $X$ to itself, i.e., $\mathfrak{S}_X = \{f:X\rightarrow X| (\forall x,y\in X~ f(x)=f(y) \implies x=y)\wedge(\forall x\in X~\exists y\in X~ f(y)=x)\}$.

(3) If $G$ is the integers mod 3 under addition and $X$ is the set $\{a,b,c,d\}$ then an example of a homomorphism $\phi: G \rightarrow \mathfrak{S}_X$ would be given by:

$\phi(0)=$ the identity function on $X$

$\phi(1)=$ gives the cycle $(abc)$ on $X$, i.e., $\phi(1)(a)=b$, $\phi(1)(b)=c$, $\phi(1)(c)=a$, and $\phi(1)(d)=d$

$\phi(2)=$ gives the cycle $(acb)$ on $X$, i.e., $\phi(2)(a)=c$, $\phi(2)(b)=a$, $\phi(2)(c)=b$, and $\phi(2)(d)=d$

Finally, you can think of $\phi$ as a homomorphism from $G \rightarrow \mathfrak{S}_X$ or as a function $G \times X \rightarrow X$ (in the latter case we write $gx$ for the output of the function and the function must meet the requirement $(hg)x=h(gx)$, which is what makes the corresponding function $G \rightarrow \mathfrak{S}_X$ a homomorphism), but NOT as a function $G \rightarrow X \times X$. You can think of it as a function $G \rightarrow {P}(X\times X)$, but only because $\mathfrak{S}_X$ is a subset of $P(X\times X)$ when you think of functions as sets of ordered pairs.

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  • $\begingroup$ I am not sure how "$ϕ(1)=$ gives the cycle $(abc)$" and likewise for 2, since it seems $1 : abcd \to bcad$ and $2 : abcd \to cabd$. $\endgroup$ – Lance Pollard Apr 21 '18 at 10:14
  • $\begingroup$ A "cycle" $(a_1 a_2 .. a_n)$ is that member of the symmetric group that maps $a_1$ to $a_2$, $a_2$ to $a_3$, ..., $a_n$ to $a_1$ and that leaves all other elements fixed. Every element of the symmetric group can be written uniquely as a product of disjoint cycles, and this is usually how mathematicians write down individual elements of the symmetric group. $\endgroup$ – C Monsour Apr 21 '18 at 10:21
  • $\begingroup$ Ah I see that makes sense now, thank you. $\endgroup$ – Lance Pollard Apr 21 '18 at 10:26
  • $\begingroup$ So in essence, $x = \varphi(a)$ returns a function from the set of bijections, and it takes as argument $a$ an element of the group. Then $y = x(b)$ takes as input a value from the set $X$ and returns a value from the set $X$... because it's a bijection! So $x$ is a bijection and $y$ is a value from the set. Ah, so the bijection can be thought of as a function, which allows $x(b)$ but it can also be thought of as a permutation. So x is a function or permutation depending on your perspective. Hopefully that's close. $\endgroup$ – Lance Pollard Apr 21 '18 at 10:42

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