Group containing no subgroup of index 2 [closed]

Solution 1:

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 $\ker(\varphi)=H$.

Solution 2:

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.