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

Statement

Suppose $G$ is a group of prime power order. Suppose $\mathcal{S}$ is a collection of subgroups of $G$. For any subgroup $H$ of $G$, denote by $n(H)$ the number of subgroups of $H$ that are in $\mathcal{S}$. Then, if $G \notin \mathcal{S}$, we have:

$n(G) \equiv \sum_{M \max G} n(M) \pmod p$

If $G \in \mathcal{S}$, we have:

$n(G) \equiv 1 + \sum_{M \max G} n(M) \pmod p$.

Related facts

Similar facts

• Congruence condition relating number of normal subgroups containing minimal normal subgroups and number of normal subgroups in the whole group
• Congruence condition on number of subgroups of given prime power order

Facts used

1. Equivalence of definitions of maximal subgroup of group of prime power order (specifically, nilpotent implies every maximal subgroup is normal, and prime power order implies nilpotent, so all maximal subgroups are normal in groups of prime power order).
2. Fourth isomorphism theorem
3. Formula for number of maximal subgroups of group of prime power order

Proof

Note that the $G \in \mathcal{S}$ case follows directly from the $G \notin \mathcal{S}$ case, so we restrict attention to the $G \notin \mathcal{S}$ case.

Given: $G$ is a group of prime power order. $\mathcal{S}$ is a collection of subgroups of $G$. For any subgroup $H$ of $G$, denote by $n(H)$ the number of subgroups of $H$ that are in $\mathcal{S}$. $G \notin \mathcal{S}$

To prove: $n(G) \equiv \sum_{M \max G} n(M) \pmod p$, where the summation is over maximal subgroups of $G$. Note that $n(G) = |\mathcal{S}|$.

Proof:

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	Let $\mathcal{T}$ be the set of ordered pairs $(H,M)$ where $H \in \mathcal{S}$ and $M$ is a maximal subgroup of $G$ containing $H$.
2	$|\mathcal{T}| = \sum_{M \max G} n(M)$.			Step (1)	[SHOW MORE] For each maximal subgroup $M$ of $G$, $n(M)$ is the number of elements of $\mathcal{T}$ whose second coordinate is $M$.
3	$|\mathcal{T}|$ is the sum, over $H \in \mathcal{S}$, of the number of maximal subgroups of $G$ containing $S$.			Step (1)	Step-direct.
4	For any subgroup $H$ of $G$, the maximal subgroups of $G$ containing $H$ are precisely the same as the maximal subgroups of $G$ containing the normal closure $H^G$ of $H$ in $G$.	Fact (1)			[SHOW MORE] By Fact (1), every maximal subgroup of $G$ is normal, so any maximal subgroup containing $H$ must contain the normal closure of $H$. Conversely, any maximal subgroup containing the normal closure of $H$ contains $H$.
5	For any subgroup $H$ of $G$, the maximal subgroups of $G$ containing $H$ correspond to maximal subgroups of $G/H^G$ where $H^G$ is the normal closure of $H$ in $G$.	Fact (2)		Step (4)	Step-fact combination direct.
6	For any proper subgroup $H$ of $G$, the number of maximal subgroups of $G$ containing $H$ is congruent to 1 mod $p$.	Fact (3)		Steps (4), (5)	[SHOW MORE] $H$ is proper in $G$ so there is at least one maximal (hence normal) subgroup of $G$ containing $H$, and in particular $H^G$ is proper, and $G/H^G$ is nontrivial. Thus, Fact (3) tells us that the number of maximal subgroups in it is congruent to 1 mod $p$.
7	$|\mathcal{T}| \equiv |\mathcal{S}| \pmod p$. Note that $|\mathcal{S}| = n(G)$.		$G \notin \mathcal{S}$	Steps (3), (6)	[SHOW MORE] By Step (3), $|\mathcal{T}|$ is a sum of $n(G)$ many counts, each of which is congruent to 1 mod $p$ by Step (6). Note that we use that all elements of $\mathcal{S}$ are proper in $G$ to apply Step (6).
8	$n(G) \equiv \sum_{M \max G} n(M) \pmod p$			Steps (2), (7)	Step-combination direct.

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}
• Counting abelian subgroups of p-groups: a projective approach by Marc Konvisser and David Jonah, Journal of Algebra, ISSN 00218693, Volume 34, Page 309 - 330(Year 1975): ^{PDF (ScienceDirect)More info}