# Equivalence of definitions of universal congruence condition

This article gives a proof/explanation of the equivalence of multiple definitions for the term collection of groups satisfying a universal congruence condition
View a complete list of pages giving proofs of equivalence of definitions

## Statement

Suppose $\mathcal{S}$ is a finite collection of finite $p$-groups, groups of prime power order for the prime $p$. The following conditions are equivalent for $\mathcal{S}$:

1. For any finite $p$-group $P$ that contains a subgroup isomorphic to an element of $\mathcal{S}$, the number of subgroups of $P$ isomorphic to elements of $\mathcal{S}$ is congruent to $1$ modulo $p$.
2. For any finite $p$-group $P$ that contains a subgroup isomorphic to an element of $\mathcal{S}$, the number of normal subgroups of $P$ isomorphic to elements of $\mathcal{S}$ is congruent to $1$ modulo $p$.
3. For any finite $p$-group $Q$ and any normal subgroup $P$ of $Q$ such that $P$ contains a subgroup isomorphic to an element of $\mathcal{S}$, the number of normal subgroups of $Q$ isomorphic to elements of $\mathcal{S}$ and contained in $P$ is congruent to $1$ modulo $p$.
4. For any finite $p$-group $P$ that contains a subgroup isomorphic to an element of $\mathcal{S}$, the number of p-core-automorphism-invariant subgroups of $P$ isomorphic to elements of $\mathcal{S}$ is congruent to $1$ modulo $P$.
5. For any finite group $G$ containing a subgroup isomorphic to an element of $\mathcal{S}$, the number of subgroups of $G$ isomorphic to an element of $\mathcal{S}$ is congruent to $1$ modulo $p$.

## Proof

### Equivalence of (1) and (2)

It suffices to prove that the number of non-normal subgroups isomorphic to elements of $\mathcal{S}$ is divisible by $p$.

Given: A finite $p$-group $P$, a collection $\mathcal{S}$ of finite $p$-groups.

To prove: The number of non-normal subgroups of $P$ isomorphic to an element of $\mathcal{S}$ is divisible by $p$.

Proof:

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 The set of non-normal subgroups of $P$ isomorphic to an element of $\mathcal{S}$ is a disjoint union of conjugacy classes of subgroups in $P$, each having size more than one. [SHOW MORE]
2 The size of any conjugacy class of non-normal subgroups in $P$ is divisible by $p$. Facts (1), (2) [SHOW MORE]
3 The number of non-normal subgroups of $P$ isomorphic to an element of $\mathcal{S}$ is divisible by $p$. Steps (1), (2) [SHOW MORE]

### Equivalence of (1) and (3)

For this, it suffices to show that the number of subgroups of $P$ isomorphic to elements of $\mathcal{S}$ and not normal inside $Q$ is divisible by $p$. This is because:

Total number of subgroups of $P$ isomorphic to elements of $\mathcal{S}$ = (Number of subgroups of $P$ isomorphic to elements of $\mathcal{S}$ that are not normal in $Q$) + (Number of subgroups of $P$ isomorphic to elements of $\mathcal{S}$ that are normal in $Q$)

Given: A finite $p$-group $P$ that is a normal subgroup of a group $Q$. A collection $\mathcal{S}$ of finite $p$-groups.

To prove: The number of subgroups of $P$ that are not normal in $Q$ and are isomorphic to elements of $\mathcal{S}$ is divisible by $p$.

Proof:

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 The set of subgroups of $P$ isomorphic to an element of $\mathcal{S}$ and not normal in $Q$ is a disjoint union of conjugacy classes of subgroups in $Q$ (the subgroups are all in $P$, but conjugacy is relative to $Q$), each having size more than one. $P$ is normal in $Q$. [SHOW MORE]
2 The size of any conjugacy class of non-normal subgroups in $Q$ is divisible by $p$. Facts (1), (2) [SHOW MORE]
3 The number of non-normal subgroups of $P$ isomorphic to an element of $\mathcal{S}$ is divisible by $p$. Steps (1), (2) [SHOW MORE]

### Equivalence of (3) and (4)

This follows by taking $Q$ as the semidirect product of $P$ and the $p$-core of $\operatorname{Aut}(P)$.

### Equivalence of (2) and (5)

Given: A finite group $G$. $p$ is a prime number. $\mathcal{S}$ is a collection of finite $p$-groups such that, in any finite $p$-group containing a subgroup isomorphic to an element of $\mathcal{S}$, the number of normal subgroups isomorphic to an element of $\mathcal{S}$ is congruent to 1 mod $p$. $G$ contains a subgroup isomorphic to an element of $\mathcal{S}$.

To prove: The number of subgroups of $G$ isomorphic to an element of $\mathcal{S}$ is congruent to 1 mod $p$.

Proof:

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 Let $P$ be a $p$-Sylow subgroup of $G$. Fact (3)
2 $P$ contains a subgroup isomorphic to an element of $\mathcal{S}$. Fact (4) $G$ contains a subgroup isomorphic to an element of $\mathcal{S}$ Step (1) [SHOW MORE]
3 The number of normal subgroups of $P$ isomorphic to an element of $\mathcal{S}$ is congruent to 1 mod $p$. Assumption of congruence condition Steps (1), (2) Step-given combination direct.
4 A subgroup of $G$ of prime power order is normalized by $P$ if and only if it is contained in $P$ and normal in $P$. Facts (2), (5) Step (1) [SHOW MORE]
5 The number of subgroups of $G$ isomorphic to an element of $\mathcal{S}$ and not normalized by $P$ is divisible by $p$. Fact (6) Step (1) [SHOW MORE]
6 The total number of subgroups of $G$ isomorphic to an element of $\mathcal{S}$ is congruent to 1 mod $p$. Steps (3), (4), (5) [SHOW MORE]