Conjugacy classes in a subgroup of symmetric group

Question

Let $H=\langle (12)(34),(456)\rangle\cong S_4$ be a subgroup of $S_6$. It is known that there are five conjugacy classes in $S_4$ and the representatives are $(1), (12), (12)(34), (123), (1234)$.

I want to find out the five representatives of conjugacy classes of $H$. If I just look at the orders of the elements, it is hard for me to determine the elements corresponding to $(12)$ and $(12)(34)$, since both two elements have order $2$ in $S_4$. Is there any way to produce the representatives without writing out all $24$ elements of $H$?

Answer 1

You have $S_4$ acting naturally on $\{3,4,5,6\}$ with odd permutations containing the cycle $(1,2)$ so the representatives $[1,(12),(12)(34),(123),(1234)]$ in the natural representation on $\{1,2,3,4\}$ become $$[1,(12)(34),(34)(56),(345),(12)(3456)]$$

Answer 2

Let $f:G\to G'$ be a isomorphism and let $\operatorname{Cl_G(a) }=\{gag^{-1}:g\in G\}$.Then $f(\operatorname{Cl_G(a) }) =\operatorname{Cl_{G^{'}}(f(a)) }$

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$H$ have exactly $5$ conjuagacy classes and corresponding representatives are $$f[(1)], f[(12) ],f[(12)(34)], f[(123)], f[(1234)]$$ where $f:S_4\to H$ is an isomorphism.