## A criterion for detecting isomorphisms among semidirect products by a cyclic group

Let be a cyclic group, an arbitrary group, and and homomorphisms such that and are conjugate subgroups in . If is infinite, assume that and are injective. Prove that by constructing an explicit isomorphism. [Hint: Suppose . Show that for some , for all . Show that the map given by is a homomorphism. Show that is bijective by constructing a two-sided inverse.]

Let .

Now , so that for some ; since is cyclic, we have for some . Now let be arbitrary in . Then . Since is arbitrary, we have for all .

We now show that is a homomorphism.

Let . Then

Thus is a homomorphism.

We now show that is bijective.

- Suppose is infinite, so that and are injective. Just as for all , there exists such that for all . Combining these results, we see that . Since is injective, , and since is infinite and arbitrary, we have . Thus .
Define by . Then , so that . Similarly, . Thus is bijective, and we have .

- Before proceeding in the finite case, we prove the following lemma, due to Luís Finotti.
Lemma: Let such that and . Then there exists such that mod and . Proof: Let and write . Now , so that ; we also write and . Let be the product of all prime divisors of which do not divide . Finally, let . Suppose is a prime divisor of . There are three cases:

- If , then since and are relatively prime. Thus does not divide .
- If and , we have two cases.
- If , then by definition. Thus . Also, since , we have . Thus .
- If , then and . Thus .

- If , then since , . Now and , so .

Since no prime divisor of divides , .

Now to the main result.

Suppose is finite. Now is cyclic of order where , by Lagrange. Since generates , generates . Since conjugation by is an isomorphism , generates . Thus . By the Lemma, there exists such that mod and . Moreover, there exists such that mod .

Define by . This map is clearly a two sided inverse of ; hence .

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## Comments

This theorem (criteria) is not difficult to prove, there is a natural way to prove the statement.

But I would like to ask the following question:

If two semi-direct products of H by K, under two homomorphisms, are isomorphic, does it imply that the images of K under these two homomorphisms in Aut(H) are conjugate?

My feeling is that this is not true in general, but I’m having a hard time coming up with a counterexample. My first idea was to let , since then has three nonconjugate subgroups of order 2 that we can use for . However, it looks like these give the quasidihedral, modular, and dihedral groups of order 16, which are not isomorphic.

This might even be true. I’ll have to think about it more.

Dear nbloomf, I had also worked same example first time, by writing presentations of these three semidirect products, and checking in GAP, what groups we get? I got same groups you mentioned. I went to higher order groups, but still I couldn’t find any counter example, and also I don’t know whether the statement I made above is true or false?

Perhaps this works.

Let be groups and group homomorphisms. Suppose is a group isomorphism. We want to show that and are conjugate in .

As a set function, for some bijections and . I claim that in fact and are isomorphisms. To see this, note that for all , we have . Comparing entries, we see that for all . So is an automorphism of . Now fix ; then and . Thus , so that is an automorphism of .

Next, we claim that conjugates into . To see this, recall the equality of first coordinates above and let and . Then for all and ; equivalently, we have for all , and since is an isomorphism, .

In particular, . Likewise, since is an automorphism of , we have , and in fact .

Eh… never mind. I’m not so sure anymore about the line . Perhaps this can be fixed, though.