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

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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.

Facts used

  1. Equivalence of definitions of maximal subgroup of group of prime power order
  2. Fourth isomorphism theorem
  3. Equivalence of definitions of size of projective space

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).

Proof details

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