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

From Groupprops
Jump to: navigation, search

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. Minimal normal implies contained in Omega-1 of center for nilpotent p-group (which follows from nilpotent implies center is normality-large)
  2. Central implies normal
  3. 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.