Conjugacy class of a subgroup of the symmetric group.

Question

If I consider $S_5$ and the subgroup $G=\langle (12),(34),(45) \rangle$ and I want to calculate the conjugacy classes of $G$. I am not sure how to do this in an efficient way. What I am doing is first explicit all the elements of $G$. $\{(12),(34),(45),(12)(34),(12)(45),(34),(354),(12)(345),(12)(354)\}$. Question 1 : Here already I am not sure I have all the elements. What would be a procedure to be sure we don't forget any? Then what I do is I go element by element : I start with $(12)$ and I look to which element it is conjugate (by multiplying $\sigma(12)\sigma^{-1}$ for each $\sigma\in G$. It seems it is only conjugate to itself. So I can get him out of the elements I want to test. Than I take $(34)$, I see it is conjugated only to itself and $(35)$ so I can take these two elements outside the the set of elements I want to test. And I continue until I have no more elements to test. Question 2 : is this the best way?

Answer 1

I think that you start to calculate too soon.

If we look at the generators of $G$ we can see $(12)$, and we can also see $(34)$ and $(45)$. Now with the swap $(12)$ we can easily get all the permutations of the set $\{1,2\}$; and using the two swaps $(34)$ and $(45)$ we can get (slightly less easy but not difficult) all the permutations of the set $\{3,4,5\}$. So what is going on here is this: $G$ is permuting $\{1,2,3,4,5\}$ in such a way that $\{1,2\}$ and $\{3,4,5\}$ don't get mixed up, and we get all such permutations. Our group $G$ is just $\text{Symm}(\{1,2\})\times\text{Symm}(\{3,4,5\})$.

Now in general when we have $G=G_1\times G_2$ it is very easy to see that $(x_1,x_2)$ and $(y_1,y_2)$ are conjugate in $G_1\times G_2$ if and only $x_1$ and $y_1$ are conjugate in $G_1$ and $x_2$ and $y_2$ are conjugate in $G_2$. So we just need to know the conjugacy classes of $\text{Symm}(\{1,2\})$ and $\text{Symm}(\{3,4,5\})$.

Now we need to calculate, but the calculations are smaller. (Or we can use the theorem that in the symmetric group elements are conjugate if and only if they have the same shape).

For $\text{Symm}(\{1,2\})$ representatives of the classes are $(1)(2)$ and $(12)$; each class has size $1$.

For $\text{Symm}(\{3,4,5\})$ representatives of the classes are $(3)(4)(5)$, $(34)(5)$, and $(345)$. The classes are $\{(3)(4)(5)\}$, $\{(34)(5), (35)(4), (45)(3)\}$, and $\{(345), (354)\}$.

So for $G$ we have $2\times 3$ classes, with representatives $(1)(2)(3)(4)(5)$, $(1)(2)(34)(5)$, $(1)(2)(345)$,$(12)(3)(4)(5)$, $(12)(34)(5)$, $(12)(345)$ of sizes $1,3,2,1,3,2$. I won't list the elements of the classes.