Showing that Klein Four Group is a normal subgroup of $S_4$

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.

• 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) – Alexey Burdin Nov 21 '19 at 23:24
• I'm advised against listing out all the possible products by my lecturer – steambuns Nov 21 '19 at 23:26
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• 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
• 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 7:31