Congruence condition on number of subgroups of given prime power order

Version for a group of prime power order
Let $$G$$ be a group of prime power order $$p^k$$ and suppose $$0 \le r \le k$$. The following are true:


 * The number of subgroups of $$G$$ of order $$p^r$$ is congruent to $$1$$ modulo $$p$$.
 * The number of normal subgroups of $$G$$ of order $$p^r$$ is congruent to 1 mod $$p$$.
 * The number of p-core-automorphism-invariant subgroups of $$G$$ of order $$p^r$$ is congruent to 1 mod $$p$$.

In particular, the collection of groups of order $$p^r$$ is a collection of groups satisfying a universal congruence condition.

Version for a general finite group
Let $$G$$ be a finite group and $$p^r$$ be a prime power dividing the order of $$G$$. Then, the number of subgroups of $$G$$ of order $$p^r$$ is congruent to 1 mod $$p$$.

Stronger facts

 * Congruence condition relating number of subgroups in maximal subgroups and number of subgroups in the whole group

Corollaries

 * Congruence condition on number of subgroups of given prime power order satisfying any given property weaker than normality
 * Congruence condition on number of normal subgroups with quotient in a specific variety in a group of prime power order

Opposite facts

 * Congruence condition fails for number of normal subgroups of given prime power order
 * Congruence condition fails for number of characteristic subgroups in group of prime power order
 * Congruence condition fails for number of central factors in group of prime power order
 * Congruence condition fails for number of transitively normal subgroups in group of prime power order

Other related facts

 * Congruence condition on number of subgroups of given prime power order and bounded exponent in abelian group: In an abelian group of prime power order, the number of subgroups of a given order and bounded exponent is either zero or congruent to one modulo $$p$$.
 * Jonah-Konvisser abelian-to-normal replacement theorem: Jonah and Konvisser establish a congruence condition on the number of abelian subgroups of order $$p^k$$ for odd primes $$p$$ and small values of $$k \le 5$$.
 * Jonah-Konvisser elementary abelian-to-normal replacement theorem: Jonah and Konvisser establish a congruence condition on the number of elementary abelian subgroups of order $$p^k$$ for odd primes $$p$$ and small values of $$k \le 5$$.
 * Equivalence of definitions of universal congruence condition
 * Congruence condition fails for subgroups of given prime power order and bounded exponent

Facts used

 * 1) uses::Congruence condition relating number of subgroups in maximal subgroups and number of subgroups in the whole group
 * 2) uses::Formula for number of maximal subgroups of group of prime power order
 * 3) uses::Congruence condition relating number of normal subgroups containing minimal normal subgroups and number of normal subgroups in the whole group
 * 4) uses::Fourth isomorphism theorem
 * 5) uses::Formula for number of minimal normal subgroups of group of prime power order

Equivalence between the multiple formulations
For a proof of the equivalence of the three formulations for a group of prime power order and the formulation for a general finite group, see collection of groups satisfying a universal congruence condition and equivalence of definitions of universal congruence condition.

Proof for a group of prime power order in terms of all subgroups using Facts (1) and (2)
The main advantage of this proof is that it has analogues for the congruence condition on number of subrings of given prime power order in nilpotent Lie ring.

This proof uses fact (1): congruence condition relating number of subgroups in maximal subgroups and number of subgroups in the whole group, combined with an induction on order.

Given: A group $$G$$ of order $$p^k$$, $$p$$ a prime number.

To prove: For any $$r \le k$$, the number of subgroups in $$G$$ of order $$p^r$$ is congruent to $$1$$ modulo $$p$$.

Proof: In this proof, we induct on $$k$$, i.e., we assume the statement is true inside groups of order $$p^l, l \le k$$.

Base case for induction: The case $$k = 0$$ is obvious.

Inductive step: If $$r = k$$, the number of subgroups is 1, so the statement is true. So we consider $$r < k$$.

For a subgroup $$H$$ of $$G$$, denote by $$n(H)$$ the number of subgroups of $$H$$ of order $$p^r$$.

Proof for a group of prime power order in terms of normal subgroups using Facts (3)-(5)
The main advantage of this proof is that it has analogues for the congruence condition on number of ideals of given prime power order in nilpotent Lie ring.

This proof uses fact (3), combined with an induction on the order.

Given: A group $$G$$ of order $$p^k$$, $$p$$ a prime number.

To prove: For any $$r \le k$$, the number of normal subgroups in $$G$$ of order $$p^r$$ is congruent to $$1$$ modulo $$p$$.

Proof: In this proof, we induct on $$k$$, i.e., we assume the statement is true inside groups of order $$p^l, l \le k$$.

Base case for induction: The case $$k = 0$$ is obvious.

Inductive step: If $$r = 0$$, the number of normal subgroups is 1, so the statement is true. So we consider $$r > 0$$.

For a subgroup $$H$$ of $$G$$, denote by $$\nu(G,H)$$ the number of normal subgroups of $$G$$ of order $$p^r$$ containing $$H$$. Denote by $$\nu(G)$$ the total number of normal subgroups of $$G$$ of order $$p^r$$.

Journal references

 * , Theorem 1.51: Hall proves the statement only for $$p$$-groups; his proof is similar to the proof presented on the wiki page.