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$G$ 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.)

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

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    $\begingroup$ What thoughts do you have? $\endgroup$ – Vladhagen Jan 27 '14 at 17:53
  • $\begingroup$ math.stackexchange.com/questions/267593/… $\endgroup$ – Vladhagen Jan 27 '14 at 17:55
  • $\begingroup$ I thought about the fact that there are only two costes and the subgroup covering the whole group. I tried to prove that the left cosets are equal to the right cosets unsuccessfully... $\endgroup$ – user124259 Jan 27 '14 at 17:57
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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 $K = H$.

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