Congruence condition on number of abelian subgroups of prime-fourth order

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Statement

Statement in terms of universal congruence condition

Let $p$ be any prime number (including the case of odd $p$ and $p=2$). Then ,the collection of abelian groups of order $p^{4}$ is a Collection of groups satisfying a universal congruence condition (?) for the prime $p$. Thus, it is also a Collection of groups satisfying a strong normal replacement condition (?) and a Collection of groups satisfying a weak normal replacement condition (?).

Hands-on statement

Let $p$ be any prime number (including the case of odd $p$ and $p=2$). Then, if $P$ is a finite $p$-group and $A$ is an abelian subgroup of $P$ of order $p^{4}$, the number of abelian subgroups of $P$ of order $p^{4}$ is congruent to 1 mod $p$.

Related facts

For a more complete list, refer collection of groups satisfying a universal congruence condition#Examples/facts.

• Congruence condition on number of abelian subgroups of prime-cube order
• Abelian-to-normal replacement theorem for prime-cube order
• Abelian-to-normal replacement theorem for prime-fourth order
• Congruence condition on number of abelian subgroups of order eight and exponent dividing four
• Congruence condition on number of abelian subgroups of order sixteen and exponent dividing eight
• Jonah-Konvisser congruence condition on number of abelian subgroups of small prime power order for odd prime
• Jonah-Konvisser congruence condition on number of elementary abelian subgroups of small prime power order for odd prime
• Congruence condition on number of abelian subgroups of small prime power order and bounded exponent for odd prime
• Congruence condition on number of cyclic subgroups of small prime power order