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

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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.

Related facts

Opposite facts

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.