TITLE: Index of a maximal subgroup in a finite group
QUESTION [12 upvotes]: My question is very simple and maybe trivial. Here it is.
Is the index of a maximal subgroup in a finite group always a prime number?
Thanks in advance.

REPLY [5 votes]: It may be worth remarking that in a finite solvable group $G$, all maximal subgroups have prime power index (there are already examples given in comments of solvable groups with maximal subgroups not of prime index). I am not sure who was the first person to prove this, but it is well-known, and can be found, for example, in D. Gorenstein's
book "Finite Groups" (its proof is elementary compared to many in that book). The proof is as follows: let $M$ be a maximal subgroup of $G$. We may suppose that $M$ contains
no nonidentity normal subgroup of $G$ (if $K$ is one such, work in $G/K$). Let $N$ be a minimal normal subgroup of $G$. Then $N$ is an elementary Abelian $q$-group for some prime $q$. Now $N$ is not contained in $M$, so $NM$ is a subgroup of $G$ which strictly contains $M$. Hence $NM = G$. Then $[G:M] = [N:N \cap M]$, and this is a power of $q$.
Note also that if $G$ is a finite solvable group with $\Phi(G) = 1$ and $L$ is any minimal normal subgroup of $G$, then there is a maximal subgroup $H$ of $G$ with 
$[G:H] = |L|$ (this is again well-known). Recall that the Frattini subgroup $\Phi(G)$ is the intersection of all maximal subgroups of $G$, and is normal. For again, $L$ is an elementary Abelian $q$-group for some prime $q$. Since $\Phi(G) = 1$, there is a maximal subgroup $H$ of $G$ which does not contain $L$. Then $G = LH$. Furthermore, $L \cap H$ is normal in $G$ (since $L$ is Abelian, $L \cap H \lhd L$, and since $L \lhd G$, we also
have $L \cap H \lhd H$). But $L$ is not contained in $H$, so $L \cap H$ is a proper subgroup of $L$. By the minimality of $L$ as a normal subgroup of $G$, we must have 
$L \cap H = 1$, so that $[G:H] = |L|$.