Congruence condition on number of subgroups of given prime power order

This article describes a congruence condition on an enumeration, or a count. It says that in a finite group and modulo prime number, the number of subgroups of given prime power order satisfies a congruence condition.
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Statement

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 $p$.

Related facts

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

Proof

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 the principle of mathematical induction in a nontrivial way (i.e., it would be hard to write the proof clearly without explicitly using induction).

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 $p^r$.

Step no.	Assertion/construction	Facts used	Given data/assumptions used	Previous steps used	Explanation
1	$n(G) \equiv \sum_{M \operatorname{max} G} n(M) \pmod p$. Here, $M \operatorname{max} G$ means that $M$ is a maximal subgroup of $G$.	Fact (1)	$r < k$		[SHOW MORE] Let $\mathcal{S}$ be the collection of subgroups of $G$ of order $p^r$. Note that, since $r < k$, $G \notin \mathcal{S}$, so Fact (1) gives the desired congruence.
2	For every $M \operatorname{max} G$, $n(M) \equiv 1 \pmod p$.		inductive hypothesis	$r < k$	[SHOW MORE] Note that $r < k$, so $r \le k - 1$, so the inductive hypothesis is applicable.
3	The number of maximal subgroups of $G$ is congruent to 1 mod $p$.	Fact (2)			Fact-direct.
4	$n(G) \equiv 1 \pmod p$			Steps (1)-(3)	[SHOW MORE] By Step (1), $n(G)$ is congruent mod $p$ to the sum $\sum_{M \operatorname{max} G} n(M)$. The latter sum is congruent mod $p$ to the sum $\sum_{M \operatorname{max} G} 1$, because of Step (2). This sum equals the number of maximal subgroups of $G$, which by Step (3) is congruent to 1 mod $p$.

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 the principle of mathematical induction in a nontrivial way (i.e., it would be hard to write the proof clearly without explicitly using induction).

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 $p^r$.

Step no.	Assertion/construction	Facts used	Given data/assumptions used	Previous steps used	Explanation
1	$\nu(G) \equiv \sum_N \nu(G,N) \pmod p$ where $N$ varies over all the minimal normal subgroups of $G$.	Fact (3)			[SHOW MORE] Let $\mathcal{S}$ be the collection of normal subgroups of order $p^r$. Then, because $r > 0$, the trivial subgroup is not in $\mathcal{S}$, so Fact (3) gives the congruence.
2	For each minimal normal subgroup $N$ of $G$, $\nu(G,N)$ is congruent to 1 mod $p$.	Fact (4)	inductive hypothesis		[SHOW MORE] $N$ has order $p$. So, normal subgroups of order $p^r$ in $G$ containing $N$ correspond, by Fact (4), to normal subgroups of order $p^r/|N| = p^{r-1}$ in $G/N$, which has order $p^{k-1}$. The inductive hypothesis now tells us that this number is congruent to 1 mod $p$.
3	The number of minimal normal subgroups of $G$ is congruent to 1 mod $p$.	Fact (5)			Fact-direct.
4	$\nu(G) \equiv 1 \pmod p$.			Steps (1), (2), (3)	[SHOW MORE] By Step (1), $\nu(G)$ is congruent to $\sum_N \nu(G,N)$ mod $p$. By Step (2), this is congruent to $\sum_N 1$, which equals the number of minimal normal subgroups of $G$. By Step (3), this number is congruent to 1 mod $p$.

References

Journal references

• A contribution to the theory of groups of prime power order by Philip Hall, Proceedings of the London Mathematical Society, ISSN 1460244X (online), ISSN 00246115 (print), Page 29 - 95(Year 1934): In this paper, Philip Hall summarizes and extends many of the known results about groups of prime power order.^{More info}, Theorem 1.51: Hall proves the statement only for $p$-groups; his proof is similar to the proof presented on the wiki page.