What a group homomorphism looks like 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:


*

*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).

*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.

*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.
 A: 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):


*

*$Id$

*$(1\ 2)=\{1\mapsto 2, 2\mapsto 1,3\mapsto 3\}$

*$(1\ 3)=\{1\mapsto 3, 2\mapsto 2,3\mapsto 1\}$

*$(2\ 3)=\{1\mapsto 1, 2\mapsto 3,3\mapsto 2\}$

*$(1\ 2\ 3)=\{1\mapsto 2, 2\mapsto 3, 3\mapsto 1\}$

*$(1\ 3\ 2)=\{1\mapsto 3, 2\mapsto 2,2\mapsto 1\}$.


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. 
A: (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.
