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

Thank you.
 A: 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 $\ker(\varphi)=H$.
A: 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. 
A: 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....
A: 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.
