$\sigma$ = (1 2 3 4 5). Find a $\gamma$ in S5 such that:

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(a) $\gamma$ $\sigma$ $\gamma^{-1}$=$\sigma^2$

(b) $\gamma$ $\sigma$ $\gamma^{-1}$=$\sigma^{-1}$

(c) $\gamma$ $\sigma$ $\gamma^{-1}$=$\sigma^{-2}$.

I thought that there was a theorem that stated $\gamma$ $\sigma$ $\gamma^{-1}$=($\gamma$(1) $\gamma$(2) $\gamma$(3) $\gamma$(4) $\gamma$(5)), so for part (a) I wound up getting $\gamma$=(3 4 5 2 1), but this is obviously wrong, so what do I do?

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Yes, the theorem works, but that would give the mapping $\gamma=(1\mapsto 3,\ 2\mapsto 5,\ 3\mapsto 2,\ 4\mapsto 4,\ 5\mapsto 1)$. (Can you write its cycle structure?) Edit: I found out that $\sigma^2$ is rather $(35241)$.

Actually, there can be other solutions as well, if you rotate the written line of the cycle, namely if you write $\sigma^2$ as e.g. $(1\ 3\ 4\ 5\ 2)$ or $(4\ 5\ 2\ 1\ 3)$ instead.