Prove that if $G$ has no subgroups of index $2$, then every subgroup of index $3$ is normal

Question

This is my question : Let G be a group without subgroups with index 2 (i.ee. there is no subgroup M s.t $[G:M]=2$).prove that every group $H<G$ s.t $[G:H]=3$ is normal in G.

What I thought to do: Let N be a subgroup s.t $[G:H]=3$ so $G/H={N,xN,yN}$ so every $g \in G$ is of the form : $g=n,g=xn,g=yn$ when $n \in \mathbb N$. I tried to prove that for every $g \in G$ : $gNg^{-1} =N$ but without a success.

Answer 1

Let $G$ act on the left coset of $H$ by left multiplication. Then we have an homomorphism $\phi$ from $G$ to $S_3$. As $G$ has no subgroup of index $2$, $\phi(G)\leq A_3$.

Notice that $\phi$ can not be trivial thus, $\phi(G)=A_3$. Then clearly $\ker \phi=H$.

Answer 2

Let $G$ act on the cosets of $H$, we obtain a morphism $G \to S_3$ with Kernel $N$. It is well known that we have $N \leq H$, we need to prove equality. If not, then we have $[G:N]= 6$ and $G \to S_3$ is surjective. By the sign we obtain a surjection $G \to S_3 \to \{-1,1\}$, hence a subgroup of index $2$.