0

I'm trying to solve a problem from my Group Theory course. It goes like this:

Being $S_n$ the symmetric group of order $n$. Find the number of index 2 subgroups of $S_n$.

I'm not pretty sure how to start. I guess it would have something to do with the normal subgroups of $S_n$, since any index two subgroup of a group $G$ is normal to $G$. But I don't know how to compute this number of subgroups, I just know that $|S_n|=n!$. Any help will be appreciated.

Alejandro Bergasa Alonso
  • 3,639
  • 1
    Just to be clear, that "duplicate" proves a stronger statement than you are asking for, at least for $n≥5$. The cases $n<5$ are "easy" to handle separately. – lulu Nov 17 '20 at 11:43
  • @lulu No, I think it doesn't. I know that any index 2 subgroup is normal, but I don't know if any normal subgroup has index two. Is this last condition true? – Alejandro Bergasa Alonso Nov 17 '20 at 11:44
  • SInce the duplicate proves that there are no normal subgroups (other than $A_n$ of course) then it instantly follows that there are no subgroups of index $2$ (other than $A_n$) As I said, the duplicate is a lot stronger than what you are asking. – lulu Nov 17 '20 at 11:47
  • In any case, here is a more exact duplicate (and it works for all $n$). – lulu Nov 17 '20 at 11:48

0 Answers0