# Formula for number of minimal normal subgroups of group of prime power order

## Statement

Suppose $G$ is a group of prime power order where the underlying prime is $p$. Suppose $s$ is the minimum size of generating set for the center $Z(G)$. Then, the number of minimal normal subgroups of $G$ equals:

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

All of these subgroups are of order $p$ and are contained in the center.

In particular, the number of such subgroups is congruent to 1 mod $p$.

## Proof

By facts (1) and (2), the minimal normal subgroups are precisely the subgroups of order $p$ contained inside $\Omega_1(Z(G))$ (the subgroup generated by elements of order $p$ in the center), which is an elementary abelian group of order $p^s$. Fact (3) now tells us that the number of such subgroups is given by the indicated formula.