# Jonah-Konvisser congruence condition on number of 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 universal congruence conditions

Suppose $p$ is an odd prime number, and $0 \le k \le 5$. Then, the set of all abelian groups of order $p^k$ (i.e., a set of representatives of all isomorphism classes of abelian groups of order $p^k$) is a Collection of groups satisfying a universal congruence condition (?). In particular, it is also a Collection of groups satisfying a strong normal replacement condition (?) and hence also a Collection of groups satisfying a weak normal replacement condition (?).

### 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 abelian subgroup of order $p^k$. The following equivalent statements hold:

1. The number of abelian subgroups of $G$ of order $p^k$ is congruent to $1$ modulo $p$.
2. The number of 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 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 abelian subgroup of order $p^k$, then $G$ has an abelian normal subgroup of order $p^k$, and moreover, $G$ has an abelian p-core-automorphism-invariant subgroup of order $p^k$.

## Related facts

### Similar congruence condition/replacement theorems

Congruence condition-cum-replacement theorem results for odd primes:

Congruence conditions for all primes:

Pure replacement theorems:

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