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

This article is about a congruence condition.
View other congruence conditions

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.

• Congruence condition on number of abelian subgroups of prime-fourth order
• Congruence condition on number of abelian subgroups of order eight and exponent dividing four
• Congruence condition on number of abelian subgroups of order sixteen and exponent dividing eight
• Abelian-to-normal replacement theorem for prime-cube order
• Abelian-to-normal replacement theorem for prime-fourth order
• Jonah-Konvisser congruence condition on number of abelian subgroups of small prime power order for odd prime: For odd $p$, the result extends to $p^{4}$ and $p^{5}$.
• Jonah-Konvisser congruence condition on number of abelian subgroups of small prime power order and bounded exponent for odd prime
• Jonah-Konvisser congruence condition on number of elementary abelian subgroups of small prime power order for odd prime

Facts used

1. Existence of abelian normal subgroups of small prime power order: This states that if $n\geq 1+k(k-1)/2$, 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 $p^{n}$, $n\geq 3$, 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. $n\geq 5$: 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 $p$.