# Jonah-Konvisser congruence condition on number of elementary abelian subgroups of small prime power order for odd prime

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This article states and (possibly) proves a fact that is true for odd-order p-groups: groups of prime power order where the underlying prime is odd. The statement is false, in general, for groups whose order is a power of two.
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## Statement

### Statement in terms of a universal congruence condition

Let $p$ be an odd prime. Suppose $0 \le k \le 5$.

Let $\mathcal{S}$ be the one-element set comprising an elementary abelian subgroup of order $p^k$. Then, $\mathcal{S}$ is a Collection of groups satisfying a universal congruence condition (?) for the prime $p$. In particular, $\mathcal{S}$ is a Collection of groups satisfying a strong normal replacement condition (?) for $p$ and hence also a Collection of groups satisfying a weak normal replacement condition (?) for $p$.

### Hands-on statement

Suppose $p$ is an odd prime number and $0 \le k \le 5$. Suppose $G$ is a finite $p$-group having an elementary abelian subgroup of order $p^k$.

The statement has the following equivalent forms:

1. The number of elementary abelian subgroups of $G$ of order $p^k$ is congruent to $1$ modulo $p$.
2. The number of elementary abelian normal subgroups of $G$ of order $p^k$ is congruent to $1$ modulo $p$.
3. If $G$ is a subgroup of a finite $p$-group $L$, then the number of elementary abelian subgroups of $G$ of order $p^k$ that are normal in $L$ is congruent to $1$ modulo $p$.

In particular, if $G$ has an elementary abelian subgroup of order $p^k$, then $G$ has an elementary abelian normal subgroup of order $p^k$. In fact, $G$ has an elementary abelian p-core-automorphism-invariant subgroup of order $p^k$, and the number of elementary abelian $p$-core-automorphism-invariant subgroups of $G$ of order $p^k$ is also congruent to $1$ modulo $p$.

### Corollary in terms of normal rank

In particular, this shows that for $p$ an odd prime and $G$ a $p$-group:

• If the rank of $G$ is less than or equal to $5$, the normal rank of $G$ is equal to the rank.
• If the normal rank is at most $4$, the rank equals the normal rank.

## Related facts

### Similar replacement theorems

For a full list of replacement theorems (including many of a completely different flavor) refer Category:Replacement theorems.

### Opposite facts

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