# Does this “Index condition” always implies that subgroup is normal?

Let $G$ be a finite group and $p$ be the smallest prime dividing $o(G)$,let $q(>p)$ be the next prime that divides $o(G)$.Suppose $G$ does not have any subgroup of index $p$ and has a subgroup say $H$ of index $q$ then $H$ is normal in $G$ ?

Edit: A simple example $A_5$ works as counterexample,So now additionally I assume that there is no prime between $p$ and $q$.

The above question is motivated from the example of $A_4$ the Alternating group on 4 symbols,since $A_4$ does not have any subgroup of index $2$ and Klein-4 group is subgroup of $G$ of index $3$ which is clearly normal.

• $\;A_5\;$ doesn't have subgroup of index $\;2,3\;$ but it has a subgroup of index $\;5\;$ , which of course isn't normal. So trying to generalize your doubt leads to a negative answer. – DonAntonio Feb 22 '16 at 15:31
• Dear @Joanpemo Thank you,what can be said if there is no prime between $p$ and $q$? – Arpit Kansal Feb 22 '16 at 15:39

Let $P$ be the image of the permutation action of $G$ on the cosets of the subgroup of index $q$. If the subgroup is not normal, then the action is not regular, so $|P| > q$. But since $|P|$ divides $q!$, $P$ must have order $p^kq$ for some $k>0$.
So $P$ is solvable by Burnside's Theorem, and hence it has a normal subgroup $N$ of order $q$. But then the inverse image of $N$ in $G$ has index $p^k$ in $G$, so $G$ has a subgroup of index $p$, contrary to assumption.