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Suppose $G$ is a finite group with normal subgroups $M,N$ and subgroups $H,K$ such that $M \cong N$, $MH=NK=G$, and $M \cap H = N \cap K = 1$. Is it the case that $H \cong K$?

Clearly $H \cong G/M$ and $K \cong G/N$, so this is similar to Isomorphic quotient groups but of course the examples there are not semi-direct products.

I assumed counterexamples would be plentiful, but unless I made a mistake, there are no examples with |G| ≤ 300.

This question was motivated by Tobias's remarks in his question How to determine if two semidirect products are isomorphic?

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Jack Schmidt
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1 Answers1

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This is just Derek Holt's answer (CW):

Let $G=C_3 \times S_3$ with normal subgroups $M=C_3 \times 1$ and $N=1 \times A_3$ and complements $H=1 \times S_3$ and $K=C_3 \times S_2$. Then $M,N$ are normal and isomorphic. $M \cap H = N \cap K = 1$ is clear, and then $MH=NK=G$ follows automatically from order considerations.

This is the unique smallest order example, and obviously generalizes to $C_p \times D_{2p}$.

In general, $$\begin{array}{rcl} G&=&(A \rtimes B) \times (A \rtimes C) \\ M&=&(A \rtimes 1) \times (1 \rtimes 1) \\ N&=&(1 \rtimes 1) \times (A \rtimes 1) \\ H&=&(1 \rtimes B) \times (A \rtimes C) \\ K&=&(A \rtimes B) \times (1 \rtimes C) \end{array}$$ is sort of the obvious thing to try, and it in fact usually works. For instance if $B$, $C$, $A \rtimes B$, and $A \rtimes C$ are all directly indecomposable, then $H \not\cong K$ as long as $B \not\cong C$. For instance Derek took $B=1$, $A=C_3$, and $C=C_2$.

Jack Schmidt
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