# Semidirect product is not left-cancellative for finite groups

## Statement

### Statement from the external semidirect product viewpoint

It is possible to have finite groups $A,B,C$, and semidirect products, such that:

$A \rtimes B \cong A \rtimes C$,

but $B$ is not isomorphic to $C$.

### Statement from the internal semidirect product viewpoint

We can construct a finite group $P$ with subgroups $H,K,L,M$ such that:

$P = H \rtimes L = K \rtimes M$,

where $H$ is isomorphic to $K$ but $L$ is not isomorphic to $M$.

## 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 top right 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 subgroup with nonzero entries in the $(34),(35),(45)$ positions is a permutable complement to $H$ in $P$. Call this subgroup $L$. Then $P$ is an internal semidirect product of $H$ and $L$. $L$ is isomorphic to the prime-cube order group:U3p.
• The subgroup with nonzero entries in the $(13),(23),(45)$ positions is a permutable complement to $K$ in $G$. Call this subgroup $M$. Then, $P$ is an internal semidirect product of $K$ and $M$. Note that $M$ is isomorphic to the elementary Abelian group of order $p^3$.
• Thus, we have $P = H \rtimes L = K \rtimes M$, with $H \cong K$ but $L$ not isomorphic to $M$.