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 $d\leq k$. 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 $p$.

Related facts

Opposite facts

• Mann's replacement theorem for subgroups of prime exponent
• Congruence condition on number of subgroups of given prime power order
• Congruence condition on Sylow numbers
• Congruence condition on number of subgroups of given prime power order and bounded exponent in abelian group

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 $p$.