Describe a natural group action

Question

$\newcommand{\Set}[1]{\left\{ #1 \right\}}$

Problem 1. For each of the following, describe a natural group action of $G$ on $X$, prove that it indeed does satisfy the necessary properties of an action, and describe the corresponding permutation representation.

I am having trouble knowing what a group action really is, how to prove it, and write its permutation. I think proving it means just to show there is an identity and an associative property?

$G = K_4$, the Klein four-group, and $X = \Set{\text{vertices of a square}}$.

Would this just be the group action of rotating the square about the axis? Please explain like I'm 5!

Answer 1

$\newcommand{\Set}[1]{\left\{ #1 \right\}}$You are doing well. Here is how you do it with permutations.

For instance, in case (i) you can write the set vertices as $X = \Set{ 0, 1, 2 }$, and map $g$ to the permutation $0 \mapsto 0$ and $1 \mapsto 2 \mapsto 1$ of $X$. ($e$ is in any case mapped to the identity.) If you know how to write permutations as products of disjoint cycles, this means $g$ is mapped to $(0) (1 2) = ( 1 2)$.

Case (ii) is similar.

For case (iii) there are two possibilities. Consider for instance the two reflections (permutations of $X$ in this case) with respect to line going through the midpoints of two opposite sides, and their product. Together with the identity, these will form a subgroup of the group of permutations of $X$, which is isomorphic to the Klein four-group.

Answer 2

By "(natural) permutation representation" of a finite group $G$ of order $n$, it is meant the isomorphic copy of $G$ in $S_n$ obtained by composing the embedding $G\stackrel{\varphi}{\hookrightarrow} S_G$ by left multiplication, after the isomorphism $S_G\to S_n$ defined by $\alpha\mapsto f^{-1}\alpha f$, where $f\colon\{1,\dots,n\}\to G$ is any bijection. For $G=K_4=\{e,a,b,ab\}$ and, for example: \begin{alignat}{1} f(1)&=e\\ f(2)&=a\\ f(3)&=b\\ f(4)&=ab\\ \tag1 \end{alignat} what is the "permutation representation" of, say, $a\in K_4$? There's no shortcut, I think, than making the full machine into operation: \begin{alignat}{1} &(f^{-1}\varphi_af)(1)=f^{-1}(\varphi_af(1))=f^{-1}(ae)=f^{-1}(a)=2\\ &(f^{-1}\varphi_af)(2)=f^{-1}(\varphi_af(2))=f^{-1}(a^2)=f^{-1}(e)=1\\ &(f^{-1}\varphi_af)(3)=f^{-1}(\varphi_af(3))=f^{-1}(ab)=4\\ &(f^{-1}\varphi_af)(4)=f^{-1}(\varphi_af(4))=f^{-1}(a^2b)=f^{-1}(b)=3\\ \end{alignat} Therefore, the "permutational alias" of $a\in K_4$ under the bijection $(1)$ is: $$(12)(34)$$ And so on with $e$ (guess what?), $b$ and $ab$. Different bijections than $(1)$ give rise to different (isomorphic) subgroups of $S_4$. And the difference is not merely set-wise: there are one copy of $K_4$ normal in $S_4$ and three nonnormal ones.