# Subgroups of index 3 are normal [duplicate]

$G$ has no subgroup with index 2. How can I show that all subgroups with index 3 are normal? (The index of the subgroup is the order of the quotient set.)

## marked as duplicate by user127.0.0.1, Jack Schmidt, Giuseppe Negro, Sharkos, TZakrevskiyJan 27 '14 at 19:05

Let $H$ be a subgroup of index $3$. Via the action of $G$ on the set $G/H$ of (left) cosets we get a group homomorphism $\alpha: G \rightarrow S_3$. Observe that $K = \ker \alpha$ is a normal subgroup of $G$ contained in $H$.
We claim that $K = H$. Note that $\alpha(G)$ does not have a subgroup of index $2$ because $G$ doesn't. Also $\alpha(G) \neq 1$ because the action of $G$ on $G/H$ is nontrivial. So $\alpha(G)$ is a nontrivial subgroup of $S_3$ which has no index $2$ subgroup. This implies that $\alpha(G) \cong C_3$ (why?) so $|G:K| = |\alpha(G)| = 3$. Hence $K = H$.