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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!

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

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  • $\begingroup$ @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

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