Dtyjlui

I came across with the following question:

Let $H={sigma in S_6 | sigma(x) textrm{ is odd if and only if } x textrm{ is odd}}$. Prove that $H$ has a normal subgroup $Y$ so $[H : Y ]=4$.

How should I prove this theorem?

I'll separate the steps into spoilers so you can use them as hints

The set $H$:

First we need to understand the set $H$, we know that a permutation in $Sym(6)$ moves points in the set $Omega:={1,2,3,4,5,6}$. Now, $H$ is defined to be the set of points which fix the subset $Lambda:={1,3,5}$ of odd integers setwise.

The group $H$:

The elements of $Sym(6)$ which fix $Lambda$ setwise are exactly the permutations that permute elements of $Lambda$, that is, $Sym(Lambda)=Sym({1,3,5})$ and the elements which permute $Gamma:={2,4,6}$, these are exactly the elements in $Sym(Gamma)$. Since no element of $Gamma$ can be sent to an element of $Lambda$ we conclude $H$ is the direct product $Sym(Lambda)times Sym(Gamma)cong Sym(3)times Sym(3)$.

The normal subgroup:

First we recall that $Sym(3)$ has a normal subgroup isomorphic to $C_3:=mathbb{Z}_3$ so we deduce that $N_1:=langle(135)rangle$ is normal in $Sym(Lambda)$ and $N_2:=langle(246)rangle$ is normal in $Sym(Gamma)$. As these are normal subgroups in different factors of the direct product we conclude there is a normal subgroup $N:=N_1times N_2$ in $H$.

The order:

A quick calculation gives $|H|=6*6=36$ and $|N|=3*3=9$. Therefore $|H:N|=4$.

Hint: for permutations of $H$, you can separate their action on odd and even integers. Thus $H cong S_3 times S_3$.