# Maximal among abelian characteristic subgroups may be multiple and isomorphic

## Statement

It is possible to have a group of prime power order $P$ with two distinct subgroups $H, K \le P$, such that both $H$ and $K$ are Maximal among abelian characteristic subgroups (?), and $H \cong K$.

## Related facts

The example described here also shows many other things:

## Proof

### Example involving the upper triangular matrices

Suppose $p$ is any prime, and let $G := U(5,p)$ be the group of upper-triangular unipotent $5 \times 5$ matrices over the field of $p$ elements. Let $P$ be the subgroup of $G$ comprising those matrices where the $(12)^{th}$ entry is zero. Then, $P$ is a group of order $p^9$.

By fact (1), we have that $G$ has two subgroups that are Abelian of maximum order: the rectangle groups of dimensions $2 \times 3$ and $3 \times 2$ respectively. Call these subgroups $H$ and $K$ respectively. Then, observe that:

• Both $H$ and $K$ are also Abelian subgroups of maximum order in $P$. Moreover, they are the only Abelian subgroups of maximum order in $P$ since they are the only Abelian subgroups of maximum order in $G$.
• $H$ and $K$ are isomorphic -- in fact, they are conjugate subgroups inside the bigger group $GL(5,p)$. This conjugation restricts to an automorphism of $G$, but not of $P$.
• Both $H$ and $K$ are normal in $G$, and hence in $P$. The quotient $P/H$ is isomorphic to the unipotent subgroup of 3-by-3 matrices while the quotient $P/K$ is isomorphic to the elementary Abelian group of order $p^3$. Hence, $H$ and $K$ are not automorphic subgroups in $P$.
• Thus, $H$ and $K$ are the only Abelian subgroups of their order, and they are not automorphic in $P$. Hence, they are both characteristic subgroups. Since they are Abelian of maximum order, they are both maximal among Abelian characteristic subgroups.

We have thus established all the required conditions.