Abelian-to-normal replacement theorem for prime-cube order

This article defines a replacement theorem
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Statement

Statement in terms of weak normal replacement condition

Let $p$ be a prime number. Then, the collection of abelian groups of order $p^{3}$ is a Collection of groups satisfying a weak normal replacement condition (?).

Hands-on statement

Suppose $p$ is a prime number and $P$ is a finite $p$-group. If $A$ is an abelian subgroup of $P$ of order $p^{3}$, there is an abelian normal subgroup $B$ of $P$ of order $p^{3}$.

Related facts

• Congruence condition on number of abelian subgroups of prime-cube order
• Congruence condition on number of abelian subgroups of prime-fourth order
• Abelian-to-normal replacement theorem for prime-fourth order
• Jonah-Konvisser congruence condition on number of abelian subgroups of small prime power order for odd prime: For $p$ odd and $0\leq k\leq 5$, the number of abelian subgroups of order $p^{k}$ is either zero or $1$ mod $p$.
• 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

Facts used

1. Existence of abelian normal subgroups of small prime power order: This states that if $n\geq 1+k(k-1)/2$, then any finite $p$-group of order $p^{n}$ has an abelian normal subgroup of order $p^{k}$.

Proof

Given: A finite $p$-group $P$ of order $p^{n},n\geq 3$, containing an abelian subgroup $A$ of order $p^{3}$.

To prove: $P$ contains an abelian normal subgroup $B$ of order $p^{3}$.

Proof: If $A=P$, then it is normal and we can set $B=A$. Thus, we assume that $A$ is a proper subgroup of $P$.

In this case, since $n\geq 4$, fact (1) tells us that $P$ has an abelian normal subgroup of order $p^{3}$, and we are done.