Conjugate in $S_n$ do not imply conjugate in its subgroup.

Question

$(12)(34)$ and $(13)(24)$ are conjugate in $S_4$, but they are not conjugate in $\mathbf V_4$, the Klein four group, since $\mathbf V_4$ is abelian.

I know that $\alpha ,\beta \in S_n , \alpha \text{ and } \beta \text{ are conjugate in $S_n$}\iff$ $\alpha$ and $\beta$ have the same number of $r$-cycles.

But I'm confused that why $(12)(34)$ and $(13)(24)$ do not have the same number of $2$-cycles in $\mathbf V_4$.

Thanks for helping.

Answer 1

The definition is that $a,b\in G$ are conjugated in $G$ if and only if there is some $g\in G$ such that $b=gag^{-1}$. In specific instances of $G$ this condition may be equivalent to other conditions which, outside of the specific context, are not equivalent to being conjugate. For instance, in an abelian group, two elements are conjugate if and only if they are equal.

Answer 2

This is a good question. The point is that two elements are conjugate in $S_{n}$ if and only if they have the same cycle structure (same number of $r$ cycles for all $r$) but this property does not apply to subgroups of $S_{n}$. Conjugation in the permutation group corresponds to a relabelling of the objects being permuted (like change of basis in vector spaces). All permutations are in $S_{n}$ so all relabellings are possible. But in subsets of $S_{n}$ some relabellings may not be available and so elements with the same cycle structure may not be conjugate.

Answer 3

For a slightly different example, the $3$-cycles (123) and (124) are (of course) conjugate in $S_4$ but not conjugate in the alternating group $A_4$, since the conjugating maps between (123) and (124) are odd permutations.