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It is given that $K_4=\{i, (1$ $2)(3$ $4), (1$ $3)(2$ $4), (1$ $4)(2$ $3)\}$. The question asks me to show that, for $h \in S_4$ and $f \in K_4$,
$$h^{-1}fh\in K_4,$$
using the order of the permutations to deduce the possible cycle-shapes of $h^{-1}fh$. I'm new to group theory so terms like isomorphic are foreign to me.

Shaun
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steambuns
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1 Answers1

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Hint: Show that the cycle structure of a permutation is preserved under conjugation.

Shaun
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  • I'm not exactly sure how to show this apart from using a few examples. Is there a formal proof for this? – steambuns Nov 21 '19 at 23:30
  • You can find a proof here and here, @Justin. – Shaun Nov 21 '19 at 23:32
  • Is this line of reasoning correct: when conjugating 2 permutations, the order follows the 1st permutation. E.g. fg follows the order of g. Following the fact that the order is 2, the possible cycle structure of $h^{-1}fh$ is is $(2^2) or (2, 1^2)$. Since $(2, 1^2)$ is not possible, the cycle structure is $(2^2)$. – steambuns Nov 21 '19 at 23:46
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    I'm afraid not, @Justin; at least not insofar as I understand what you're saying. I'm not sure what "follows the order of" means. Regardless: what you need to show is that, $$\begin{align}h^{-1}fh&=h^{-1}(ab)(cd)h \ &=(h(a)h(b))(h(c)h(d)),\end{align}$$ where $h(x)$ is $h$ applied to the underlying element $x\in{1,2,3,4}$. – Shaun Nov 21 '19 at 23:53
  • @Shaun how do you get from the first to the second line in: $$\begin{align}h^{-1}fh&=h^{-1}(ab)(cd)h \ &=(h(a)h(b))(h(c)h(d)),\end{align}$$ – baked goods Nov 22 '19 at 07:31
  • I've linked to proofs in this comment, @panzershreks. – Shaun Nov 22 '19 at 07:50