# Congruence condition on number of abelian subgroups of prime-cube order

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## Statement

### Statement in terms of universal congruence condition

Let $p$ be a prime number (odd or $p = 2$). The collection of abelian $p$-groups of order $p^3$ is a Collection of groups satisfying a universal congruence condition (?). Thus, it is also a Collection of groups satisfying a strong normal replacement condition (?) and a Collection of groups satisfying a weak normal replacement condition (?).

### Hands-on statement

Let $p$ be any prime (including the cases of odd $p$ and $p = 2$). Then, if $P$ is a finite $p$-group and $A$ is an abelian subgroup of $P$ of order $p^3$, the number of abelian subgroups of $P$ of order $p^3$ is congruent to $1$ modulo $p$.

## Related facts

For a summary of facts about universal congruence conditions, refer collection of groups satisfying a universal congruence condition#Examples/facts.

## Facts used

1. Existence of abelian normal subgroups of small prime power order: This states that if $n \ge 1 + k(k-1)/2$, then any group of order $p^n$ contains an abelian normal subgroup of order $p^k$.
2. Congruence condition on number of abelian subgroups of prime index
3. Congruence condition relating number of subgroups in maximal subgroups and number of subgroups in the whole group: This states that if $n(G)$ is the number of subgroups of $G$ isomorphic to one of the groups in a given collection $\mathcal{S}$ and $G \notin \mathcal{S}$, then $n(G) \equiv \sum n(M) \pmod p$.
4. Congruence condition on number of subgroups of given prime power order

## Proof

Given: A group $G$ of order $p^n$, $n \ge 3$, that contains an abelian subgroup of order $p^3$.

To prove: The number of abelian subgroups of $G$ of order $p^3$ is congruent to $1$ modulo $p$.

Proof: We consider three cases:

1. $n = 3$: In this case, there is exactly one abelian subgroup of order $p^3$, namely $G$ itself.
2. $n = 4$: In this case, fact (2) completes the proof.
3. $n \ge 5$: In this case, we use induction on $n$. By fact (1), every maximal subgroup of $G$ contains an abelian normal subgroup (normality being relative to the maximal subgroup) of order $p^3$, so by the inductive hypothesis, the number of abelian subgroups of order $p^3$ contained in each maximal subgroup is congruent to $1$ modulo $p$. Since the number of maximal subgroups is itself congruent to $1$ modulo $p$ by fact (4), fact (3) tells us that the number of abelian subgroups of $G$ of order $p^3$ is congruent to $1$ modulo $p$.