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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.

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  • $\begingroup$ At least you can do it manually or with a computer, since $|S_4|=24$ and you know how to get product of two permutations. (it is not meant to be a clue for solve) $\endgroup$ – Alexey Burdin Nov 21 '19 at 23:24
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    $\begingroup$ I'm advised against listing out all the possible products by my lecturer $\endgroup$ – steambuns Nov 21 '19 at 23:26
  • $\begingroup$ After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $\checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?. $\endgroup$ – Shaun Nov 22 '19 at 1:49
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Hint: Show that the cycle structure of a permutation is preserved under conjugation.

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  • $\begingroup$ I'm not exactly sure how to show this apart from using a few examples. Is there a formal proof for this? $\endgroup$ – steambuns Nov 21 '19 at 23:30
  • $\begingroup$ You can find a proof here and here, @Justin. $\endgroup$ – Shaun Nov 21 '19 at 23:32
  • $\begingroup$ 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 $(2^2)$. $\endgroup$ – steambuns Nov 21 '19 at 23:46
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    $\begingroup$ 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 $x\in\{1,2,3,4\}$. $\endgroup$ – Shaun Nov 21 '19 at 23:53
  • $\begingroup$ @Shaun how do you get from the first to the second line in: $$\begin{align}h^{-1}fh&=h^{-1}(ab)(cd)h \\ &=(h(a)h(b))(h(c)h(d)),\end{align}$$ $\endgroup$ – baked goods Nov 22 '19 at 7:31

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