Congruence condition fails for subgroups of order p^p and exponent p
From Groupprops
Contents |
Statement
Let p be a prime number. Suppose P is a group of prime power order for the prime p. Let pk be a prime power less than or equal to the order of P and pd be a power of p with
. It is not necessary that the number of subgroups of order pk and exponent dividing pd 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 22 = 4 and exponent 21 = 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 pp + 1 and exponent p2. Consider the subgroups of P having order pp and exponent p. There are exactly two of these, both of them isomorph-free: the elementary abelian normal subgroup of order pp, and the semidirect product of the commutator subgroup (which has order pp − 1 with the wreathing element of order p.