Let V be a finite dimensional vector space over the field with p elements, where p is a prime number. Given arbitrary α ,β ∈ GL(V) , we consider the semidirect products V α and V β , and show that if V α and V β are isomorphic, then α must be similar to a power of β that generates the same subgroup as β ; that is, if H and K are cyclic subgroups of GL(V) such that V H V K , then H and K must be conjugate subgroups of GL(V) . If we remove the cyclic condition, there exist examples of nonisomorphic , let alone nonconjugate, subgroups H and K of GL(V) such that V H V K . Even if we require that noncyclic subgroups H and K of GL(V) be abelian, we may still have V H V K with H and K nonconjugate in GL(V) , but in this case, H and K must at least be isomorphic. If we replace V by a free module U over Z/pᵐZ of finite rank, with $m>1$ , it may happen that U H U K for nonconjugate cyclic subgroups of GL(U) . If we completely abandon our requirements on V , a sufficient criterion is given for a finite group G to admit nonconjugate cyclic subgroups H and K of Aut(G) such that G H G K . This criterion is satisfied by many groups.
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Gebhardt et al. (2024) studied this question.
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