$G$, a subgroup of $S_6$ has an element of order $6$. Prove that $G$ has a normal subgroup $H$ of index 2.

Question

So I know this subgroup G must have either a 6-cycle or a 2-cycle times a 3-cycle as an element.

I want to prove that $G$ has a normal subgroup $H$ of index 2, which is the same thing as saying I want to prove that G has a subgroup H where H has half the elements in $G$. Where do I go next?

Answer 1

Since $G$ has an element $g$ of order $6$ it means it contains an element that is a product of odd number of transpositions. Also $G$ has an element that is a product of even number of transposition, namely $g^2$.

Now consider the homomorphism $\sigma: G \to \{-1,1\}$, given by $\sigma(g) = 1$ if $G$ can be written as even number of trasposition and $\sigma(g) = -1$, otherwise. The first paragraph shows us that the homomorphism is surjective. Hence by the first Isomorphism Theorem $\frac G{\ker \sigma} \equiv \{-1,1\}$, i.e $[G:\ker \sigma] = 2$

NOTE The group action acting on $\{-1,1\}$ is the usual multiplication. In fact this group is isomorphic to $Z_2$ under addition.