# Congruence condition fails for subgroups of order p^p and exponent p

## Statement

Let $p$ be a prime number. Suppose $P$ is a group of prime power order for the prime $p$. Let $p^k$ be a prime power less than or equal to the order of $P$ and $p^d$ be a power of $p$ with $d \le k$. It is not necessary that the number of subgroups of order $p^k$ and exponent dividing $p^d$ is either zero or congruent to $1$ modulo $p$.

## Proof

### The case $p = 2$

Consider the dihedral group of order eight. The number of subgroups of order $2^2 = 4$ and exponent $2^1 = 2$ is equal to $2$, which is neither equal to zero or congruent to $1$ modulo $2$.

### The case of odd $p$

Let $p$ be an odd prime, and let $P$ be the wreath product of groups of order p. Then, $P$ is a group of order $p^{p+1}$ and exponent $p^2$. Consider the subgroups of $P$ having order $p^p$ and exponent $p$. There are exactly two of these, both of them isomorph-free: the elementary abelian normal subgroup of order $p^p$, and the semidirect product of the commutator subgroup (which has order $p^{p-1}$) with the wreathing element of order $p$.
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