Group action question

Question

Take $G=\mathbb{Z_6}$ and $X=\{a,b,c,d,e\}$.

This is a group action that my teacher did during class and I'm not understanding this example, please help me understand!!

Define the action as follows:

$1 \cdot a =b$

$1 \cdot b=c$

$1 \cdot c=a$

$1 \cdot d=e$

$1 \cdot e=d$

Why aren't we defining the group action for the other elements in $\mathbb{Z_6}$ like $2,3,4,5$? Or is my teacher saying that this is sufficient to define a group action?

My teacher then said we must check some things:

Think of it as a homomorphism $\rho: \mathbb{Z_6} \to S_{\{a,b,c,d,e\}}$ where $\rho(1)=(abc)(de)$ i'm not sure why or how this is a homomorphism but I do understand that $\rho(1)=(abc)(de)$ though.

And we decided to check the orders of both sides, $|(abc)(de)|=lcm(3,2)=6=|1|$ I understand this computation but I don't know why we are calculating this.

Answer 1

I agree that is this is a strange way to present a group action, but let’s puzzle it out. First, we need to know what we’re even talking about. Let $G$ be a group and $X$ be a set. A (left) group action of $G$ on $X$ is a function

$$(~\!\cdot~\!): G\times X \to X$$

where I’m going to denote $(~\!\cdot~\!)(g,x)$ by $x^g$. We want this function to have two properties:

1. if $e$ is the identity in $G$, then $x^e = x$ for all $x\in X$, and
2. if $g$ and $h$ are elements of $G$, then $(x^g)^h = x^{hg}$ for all $x\in X$.

Now, notice that for each $g\in G$ I can create a function

$$\square^g: X\to X$$

where we define $\square^g(x) = x^g$. Moreover, since $g$ has to have an inverse, we know that $\square^g$ is actually a bijection, or a permutation, on $X$! You should check a couple of things here:

1. The set of all bijections on $X$, denoted by $\operatorname{Sym}(X)$, forms a group under function composition.
2. There is a well-defined function $\rho:G\to \operatorname{Sym}(X)$ where $\rho(g) = \square^g$.
3. This $\rho$ function is actually a group homomorphism! To prove this, use the two facts about group actions above.

---

You’re exactly correct when you saying that the information your teacher gave you is just sufficient to find all the rest of the details. For example, if I want to find $b^3$, I can say that

$$b^3 = b^{2+1} = (b^1)^2 = c^2 = c^{1+1} = (c^1)^1 = a^1 = b$$

and so $b^3 = b$. I can do this because $1$ is a generator of $\mathbb{Z}/6\mathbb{Z}$. This is also why we calculate the order of both $1$ and of $(abc)(de)$. The fact the both orders are $6$ tells us that $\rho$ is actually injective. Why?