# Congruence condition on number of abelian subgroups of small prime power order and bounded exponent for odd prime

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## Statement

### In terms of universal congruence condition

Suppose $p$ is an odd prime, and $0 \le d \le k \le 5$. Then, the collection of abelian groups of order $p^k$ and exponent dividing $p^d$ is a Collection of groups satisfying a universal congruence condition (?).

### Hands-on statement

Suppose $p$ is an odd prime, and $P$ is a finite $p$-group. Suppose $0 \le d \le k \le 5$. Suppose $P$ has an abelian subgroup of order $p^k$ and exponent dividing $p^d$. Then, the following equivalent statements hold:

1. The number of abelian subgroups of $P$ of order $p^k$ and exponent dividing $p^d$ is either equal to zero or congruent to $1$ modulo $p$.
2. The number of abelian normal subgroups of $P$ of order $p^k$ and exponent dividing $p^d$ is congruent to $1$ modulo $p$.
3. For any finite $p$-group $L$ containing $P$, the number of abelian subgroups of $P$ of order $p^k$ and exponent dividing $p^d$ that are normal in $L$ is congruent to $1$ modulo $p$.