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 $$p$$.

Facts used

 * 1) uses::Minimal normal implies contained in Omega-1 of center for nilpotent p-group (which follows from nilpotent implies center is normality-large)
 * 2) uses::Central implies normal
 * 3) uses::Equivalence of definitions of size of projective space

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.