I solved the following question: given the subgroup of $S_3\times S_3$
$$G_3 := \left\{ (x,y) \in S_3\times S_3 : x\equiv y \bmod A_3 \right\},$$
find all it's conjugacy classes and their corresponding orders. I proceeded like this, using, where necessary, the fact that conjugation preserves cycle type:
- $G_3$ is the set of pairs of permutations of the same type: either both are even or both are odd. This allow us to write
$$G_3 = [(12)A_3\times (12)A_3]\sqcup[A_3\times A_3],$$
where $\sqcup$ denotes disjoint union.
With that, we can list the conjugacy classes:
we have the class of $(1,1)$, with only 1 element;
we have the class of elements of the type $((**),(\star\star))$, i.e., pairs of transpositions, which has 9 elements since only $1$ and the own pair of transpositions stabilizes this kind of element;
we have the class of the elements of the type $(1,(***))$, which has only 2 elements since every conjugate of 1 is 1, the conjugation of a 3-cycle by a transposition gives the other 3-cycle (which is the inverse of the original) and the conjugation of a 3-cycle by a 3-cycle is the original 3-cycle;
we have the classes of the elements of the type $((***),1)$, $((***),(***))$ and $((***),(\star\star\star))$ with 2 elements each for the same reasons as presented on the previous item.
My question is wether this approach can be generalized for $S_n\times S_n$. Namely, if
$$G_n = \left\{ (x,y) \in S_n\times S_n : x\equiv y \bmod A_n \right\},$$
can we follow the same line as with $n=3$ to find all the conjugacy classes of $G_n$?
What I (think) I have found so far is that we can still write
$$[(12)A_n\times (12)A_n]\sqcup[A_n\times A_n]$$
since we are still thinking of pairs of permutations of the same parity. But how would the previous method transport for this case?
PS: by $x\equiv y\bmod A_n$ I want to mean $xy^{-1} \in A_n$.
PSS: this is my first posted question, so I apologize in advance for any mistakes or inadequate formatting. :-)