# Formula for number of maximal subgroups of group of prime power order

## Statement

Suppose $G$ is a group of prime power order where the underlying prime is $p$. Suppose $r$ is the minimum size of generating set for $G$. Then, the number of maximal subgroups of $G$ is equal to

$\frac{p^r - 1}{p - 1} = p^{r-1} + p^{r-2} + \dots + p + 1$

In particular, for $r \ge 1$, this number is congruent to 1 mod $p$.

## Proof

### Proof outline

The key idea behind the proof is that the maximal subgroups of $G$ correspond, via the fourth isomorphism theorem, to maximal subgroups of the Frattini quotient $G/\Phi(G)$. The latter is an elementary abelian group of order $p^r$ and the number of maximal subgroups equals the number of codimension one subspaces, which is the indicated value by Fact (3).