Let $G=(\Bbb Z/12\Bbb Z)^\times$ and $S=(\Bbb Z/13\Bbb Z)^\times$. Show that $\alpha$ is a group action of $G$ on $S$.

Question

I have the following exercise:

Let $G=(\Bbb Z/12\Bbb Z)^\times$

1. Write down the elements of $G$. Is $G$ cyclic?

The solution manual says: $G=\{1,5,7,11\}$

My thoughts: I think i get why $5, 7, 11$ are elements. However I'm not sure if $1$ is always an element, or why $1$ is an element.

Furthermore the solution manual says: $G$ is not cyclic since no element has order $4$.

My thoughts: So why are we interested if they have order $4$? Is it because there are $4$ elements?

---

So anyway the next part of the exercise is where I really get into trouble.

Exercise continued:

Let now $S=(\Bbb Z/13\Bbb Z)^\times$. We define the map $\alpha:G\times S\rightarrow S$ by $\alpha([a]_{12},[s]_{13})=[a]_{12}\cdot[s]_{13}=[s^a]_{13}$

2. Show that alpha is a group action of $G$ on $S.$

My thoughts: So I don't understand the text in the exercise, and don't know how to start on question (2). Can anyone help?

Thank you!

Answer 1

$G$ is the multiplicative group of $\Bbb Z/12\Bbb Z$ whose elements are the classes $\bar a$ such that ${\rm gcd}(a,12)=1$. Thus $$ G=\{\bar1,\bar5,\bar7, \bar{11}\}. $$ A group is cyclic, by definition, if there's an element $g$ in it whose powers exhaust all the element of the group. For a group with $n$ elements to be cyclic is necessary and sufficient that there exist an element of exact order $n$.

This is not the case for the $G$ in question since $$ \bar1^2=\bar5^2=\bar7^2=\bar{11}^2=\bar1, $$ and so there are no elements of order $4$.

An action of a group $G$ on a set $S$ is a function $G\times S\rightarrow S$ satisfying some requirements (which I will not recall here). The main point of the exercise under scrutiny is that before checking the action axioms one has to make sure that the map in question is well defined. Since the definition $[a]_{12}\cdot[s]_{13}=[s^a]_{13}$ involves the choice of a representant of the class $[a]_{12}$ this means that the result must be shown independent of this choice, namely one has to show that $$ \text{if $[a]_{12}=[b]_{12}$ then $[s^a]_{13}=[s^b]_{13}$.} $$ For, as observed already by S. Dolan in his answer, Fermat's theorem turns out useful.

I leave the details (and the rest of the problem) to you.

Answer 2

Part 1

$1$ is an element of $G$ because it is coprime to $12$. Indeed, its inclusion is essential since it is the identity element.

Part 2

The group action on $S$ (if I've interpreted your symbols correctly) is that the element $a$ of $G$ maps $s\in S$ to $s^a$.

You should be able to check the conditions that you have been given for a group action (identity and compatibility) but ask if you get stuck. However, it might be worth knowing that the reason this works is because, by Fermat's little theorem, $s^{12}\equiv 1$ modulo $13$.

Answer 3

Thank you! I think I godt it, but this group action is definitley not my favorit topic, so I get stuck in the next exercises as well.

(3) Determine the orbit $G⋅[2]_{13}$ of $[2]_{13}$ in $S$

(4) Determine the stabilizer $G_{[3]_{13}}$ of $[3]_{13}$ in $S$

I've read and watched videos about orbit and stabilizer, but can't seem to convert it to this example.