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.

  • $\begingroup$ The group action of $2, 3, 4,$ and $ 5$ is given from the action of $1$ and the compatibility of group action with group operation (addition modulo $6$); e.g., $2\cdot a=1\cdot(1\cdot a)$ since $2=1+1$ $\endgroup$ – J. W. Tanner Jul 22 at 19:46
  • $\begingroup$ Since $\mathbb{Z}_6$ is cyclic, homomorphisms from $\mathbb{Z}_6\to S_\{a,b,c,d,e\}$ are determined by the image of a generator (in this case, $1$). Does that help? $\endgroup$ – rogerl Jul 22 at 19:50
  • $\begingroup$ Could you be more explicit @rogerl? I would like to clear this understanding more explicitly there's been so many situations like these ones just don't get it. $\endgroup$ – Skypanties Jul 22 at 19:56
  • $\begingroup$ And @J.W.Tanner I figured that part out thank you. $\endgroup$ – Skypanties Jul 22 at 19:57
  • 2
    $\begingroup$ @DonAntonio Presumably 1 is used as the number and the group is additive. $\endgroup$ – Tobias Kildetoft Jul 22 at 21:02

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?


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