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Prove that if a group is containing no subgroup of index 2 then any subgroup of index 3 is normal.

Thank you.

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What have tried? You added some questions here without showing others any attempts!! – Nancy Rutkowskie Dec 30 '12 at 13:56

Sketch of proof: Let $H\le G$ of index $3$. Denote $X=G/H$ and consider the map $\varphi:G\to\operatorname{Sym}(X)\cong S_3$ defined by $[\varphi(g)](Hx)=Hxg^{-1}$.
1) Show that $\varphi$ is a group homomorphism.
2) Show that $\ker(\varphi)\subseteq H$.
3) Using the first isomorphism theorem, deduce that $\ker(\varphi)$ is of index $3$ in $G$.
4) Deduce that $\ker(\varphi)=H$.

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I'm not sure what is going on here, but the existence of the subgroup is assumed ... the result does not apply in the case where there is no subgroup of the requisite index. – Mark Bennet Dec 30 '12 at 13:49
@MarkBennet: if there is no subgroup of index 3, then it is certainly true that every subgroup of index 3 is normal! More than that, they're all central! ;) – tomasz Dec 30 '12 at 13:58
@tomasz - I'm not understanding the problem with $A_4$ ... – Mark Bennet Dec 30 '12 at 13:59
@MarkBennet: those comments concerned the previous version of this answer. I will delete them now. – tomasz Dec 30 '12 at 14:00
@DennisGulko: Dear Dennis, I refereed your answer to this question. Have a look plz. – Babak S. May 21 '13 at 16:40

Suppose $G$ is a finite group with no subgroup of index $2$. Let $H$ be a subgroup of index $3$. In that case, there is a normal subgroup K contained in H, such that $[G : K]$ divides $3$. Since $G$ has no subgroup of index $2$, so $[G : K] = 1, 3, 6.$

If $G/K$ ~ $S_3$, then $G/K$ contains a subgroup $H/K$ of index $2$, since $S_3$ does; but now the correspondence theorem gives

$[G/K : H/K] = [G : H] = 2$,

contrary to assumption

Hence $|K| = |H| ==> K = H$, since $K$ is contained in $H$, therefore $H$ is normal.

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I think you have to prove the existence of $K$ as a normal subgroup. – Mark Bennet Dec 30 '12 at 13:51

We know that:

If $G$ is a group such that for subgroup, say $H$, $[G:H]=n<\infty$ then there is a normal subgroup, say $K$, in $G$ such that $K\leq H$ and $[G:k]$ is finite and divides $n!$.

For proof the above fact, you can use the way @Dennis noted in brief and so you can build your own proof.

Hence, here we have such $K$ with $[G:K]\big|3!=1\times 2\times 3$. Obviously, $[G:K]\neq 1, \neq2$ so....

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Suppose that $H$ is a subgroup of index $3$. According to Theorem 4.4 of Basic abstract Algebra of BHATTACHARYA, we know that there is a homomorphism $‎\varphi‎:G ‎\rightarrow‎ S_3$ such that $Ker(‎\varphi‎)=‎\cap‎ xHx^{-1}$. So $|Img(\varphi) |\mid 6$ and $Ker(\varphi)\subset H$. Therefore $|Img(\varphi)|\in \{1, 2, 3, 6\}$ and $[G:Ker(\varphi)]\geq 3$. But we know that $G/Ker(\varphi) \simeq Img(\varphi)$, then $|Img(\varphi)|\in \{ 3, 6\}$. If $|Img(\varphi)|=6$, then we have $G/Ker(\varphi)\simeq S_3$. But in $S_3$ we have $H=\{ I, (1 2 3), (1 3 2)\}$ is normal and of index $2$. So there exist a subgroup of index $2$ in $G$, a contradiction. So $|Img(\varphi)|=3$ and therefore $|G/Ker(\varphi)|=3$ and since $Ker(\varphi)\subset H$ we conclude that $Ker(\varphi)=H$ and then $H$ is normal.

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