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

This article is about a congruence condition.

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## Contents

## Statement

### Statement in terms of universal congruence condition

Let be a prime number (odd or ). The collection of abelian -groups of order 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 be any prime (including the cases of odd and ). Then, if is a finite -group and is an abelian subgroup of of order , the number of abelian subgroups of of order is congruent to modulo .

## 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 , the result extends to and .
- 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

- Existence of abelian normal subgroups of small prime power order: This states that if , then any group of order contains an abelian normal subgroup of order .
- Congruence condition on number of abelian subgroups of prime index
- Congruence condition relating number of subgroups in maximal subgroups and number of subgroups in the whole group: This states that if is the number of subgroups of isomorphic to one of the groups in a given collection and , then .
- Congruence condition on number of subgroups of given prime power order

## Proof

**Given**: A group of order , , that contains an abelian subgroup of order .

**To prove**: The number of abelian subgroups of of order is congruent to modulo .

**Proof**: We consider three cases:

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