Permutations that conjugate in $S_5$ but not in $A_5$

Question

Find two permutations that conjugate in $S_5$, but not in $A_5$.

I can't understand why is it possible - in order for two permutations to conjugate, they must have the same cycle structure.

If two permutations are conjugate in $S_5$, this means they have the same cycle structure, and therefore will have the same structure in $A_5$, and will be still conjugate in $A_5$...

What am I missing?

Answer 1

As you said, being conjugate in $S_5$ is equivalent to having the same cycle structure. But this is not true in $A_5$. Two permutations that are conjugate in $A_5$ will have the same cycle structure, but the converse is not necessarily true. $x,y \in A_5$ and $y=gxg^{-1}$ where $g \in S_5 \backslash A_5$ then it implies $x$ and $y$ are conjugate in $S^5$ but they might not be conjugate in $A^5$.

Answer 2

If $\alpha$ is a $5$-cycle in $S_5$ then the permutations in $S_5$ which commute with it are the powers of $\alpha$ which are all even. If $\beta$ is an odd permutation, and $\alpha'=\beta\alpha\beta^{-1}$ then $\alpha'$ is a conjugate of $\alpha$ in $S_5$ but not in $A_5$. For if $\alpha'=\beta'\alpha\beta'^{-1}$ for $\beta'\in A_5$ then $\beta^{-1}\beta'$ commutes with $\alpha$, so lies in $A_5$. That forces $\beta\in A_5$, a contradiction.

I like this geometric picture. $A_5$ is isomorphic to the rotation group of the regular icosahedron. That has a rotation of order $5$ with angle $2\pi/5$. Its square is a rotation of order $5$ with angle $4\pi/5$. These two rotations cannot be conjugate; these correspond to non-conjugate $5$-cycles in $A_5$.