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

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!

  • $\begingroup$ So ex = x, gv1 = v1, gv2 = v2, and g*v2 = v3? Is that correct and what is the "permutation" $\endgroup$ – 7th Guy Apr 23 '18 at 11:34

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

| cite | improve this answer | |
  • $\begingroup$ @Servaes misprint fixed, thanks. $\endgroup$ – Andreas Caranti Apr 23 '18 at 12:28
  • $\begingroup$ I still do not understand what the "action" G takes on to X is in i and ii $\endgroup$ – 7th Guy Apr 23 '18 at 17:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.