# Abelian-to-normal replacement theorem for prime-square index

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## Statement

Let $p$ be any prime number (possibly $p = 2$). Then, if $P$ is a finite $p$-group (i.e., a group of prime power order where the underlying prime is $p$) and $A$ is an abelian subgroup of index $p^2$ in $P$, then there is an abelian normal subgroup of $P$ of index $p^2$. Moreover, we can choose this abelian normal subgroup so that it is contained in the normal closure of $A$ in $P$.

## Facts used

1. Congruence condition on number of abelian subgroups of prime index

## Proof

Given: A finite $p$-group $P$ for some prime $p$, an abelian subgroup $A$ of $P$ of index $p^2$.

To prove: There exists an abelian normal subgroup $B$ of $P$ of index $p^2$, contained inside the normal closure of $A$ in $P$.

Proof:

1. If $A$ is normal in $P$, we are done. Otherwise, $A$ is a 2-subnormal subgroup and its normal closure is a maximal subgroup $M$ of $P$ containing $A$. $M$ is normal and has index $p$ in $P$, and $A$ is normal and has index $p$ in $M$.
2. By fact (1), the number of abelian subgroups of $M$ is congruent to $1$ modulo $p$.
3. Since $M$ is normal in $P$, $P$ acts on $M$ by conjugation and the orbits of abelian maximal subgroups are therefore of size equal to a power of $p$. Since the total number of abelian maximal subgroups is $1$ modulo $p$, there exists an orbit of size $1$, and the member of this orbit is thus an abelian normal subgroup of index $p$.