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

Let K be a cyclic group, H an arbitrary group, and \varphi_1 and \varphi_2 homomorphisms K \rightarrow \mathsf{Aut}(H) such that \mathsf{im}\ \varphi_1 and \mathsf{im}\ \varphi_2 are conjugate subgroups in \mathsf{Aut}(H). If K is infinite, assume that \varphi_1 and \varphi_2 are injective. Prove that H \rtimes_{\varphi_1} K \cong H \rtimes_{\varphi_2} K by constructing an explicit isomorphism. [Hint: Suppose \sigma \varphi_1[K] \sigma^{-1} = \varphi_2[K]. Show that for some a \in \mathbb{Z}, \sigma \varphi_1(k) \sigma^{-1} = \varphi_2(k)^a for all k \in K. Show that the map \psi : H \rtimes_{\varphi_1} K \rightarrow H \rtimes_{\varphi_2} K given by \psi((h,k)) = (\sigma(h), k^a) is a homomorphism. Show that \psi is bijective by constructing a two-sided inverse.]

Let K = \langle x \rangle.

Now \sigma \varphi_1(x) \sigma^{-1} \in \varphi_2[K], so that \sigma \varphi_1(x) \sigma^{-1} = \varphi_2(y) for some y; since K is cyclic, we have y = x^a for some a. Now let k = x^b be arbitrary in K. Then \sigma \varphi_1(k) \sigma^{-1} = \sigma \varphi_1(x)^b \sigma^{-1} = (\sigma \varphi_1(x) \sigma^{-1})^b = \varphi_2(x)^{ab} = \varphi_2(k)^a. Since k is arbitrary, we have \sigma \varphi_1(k) \sigma^{-1} = \sigma_2(k)^a for all k \in K.

We now show that \psi is a homomorphism.

Let (h_1,k_1),(h_2,k_2) \in H \rtimes_{\varphi_1} K. Then

\psi((h_1,k_1)(h_2,k_2)) = \psi((h_1 \varphi_1(k_1)(h_2), k_1k_2))
= (\sigma(h_1 \varphi_1(k_1)(h_2)), (k_1k_2)^a)
= (\sigma(h_1) \sigma(\varphi_1(k_1)(h_2)), k_1^a k_2^a)
= (\sigma(h_1) (\sigma \circ \varphi_1(k_1))(h_2), k_1^a k_2^a)
= (\sigma(h_1) (\varphi_2(k_1)^a \circ \sigma)(h_2), k_1^a k_2^a)
= (\sigma(h_1) \varphi_2(k_1^a)(\sigma(h_2)), k_1^a k_2^a)
= (\sigma(h_1), k_1^a)(\sigma(h_2), k_2^a)
= \psi((h_1,k_1)) \psi((h_2,k_2)).

Thus \psi is a homomorphism.

We now show that \psi is bijective.

  1. Suppose K is infinite, so that \varphi_1 and \varphi_2 are injective. Just as \sigma \varphi_1(k) \sigma^{-1} = \varphi_2(k)^a for all k, there exists b such that \sigma^{-1} \varphi_2(k) \sigma = \varphi_1(k)^b for all k. Combining these results, we see that \varphi_2(k) = \varphi_2(k^{ab}). Since \varphi_2 is injective, k^{1-ab} = 1, and since K is infinite and k arbitrary, we have ab = 1. Thus a = b \in \{1,-1\}.

    Define \chi : H \rtimes_{\varphi_2} K \rightarrow H \rtimes_{\varphi_1} K by \chi((h,k)) = (\sigma^{-1}(h), k^a). Then (\chi \circ \psi)((h,k)) = \chi(\psi((h,k))) = \chi(\sigma(h), k^a) = ((\sigma^{1-}\sigma)(h), k^{aa}) = (h,k), so that \chi \circ \psi = 1. Similarly, \psi \circ \chi = 1. Thus \psi is bijective, and we have H \rtimes_{\varphi_2} K \cong H \rtimes_{\varphi_1} K.

  2. Before proceeding in the finite case, we prove the following lemma, due to Luís Finotti.

    Lemma: Let a,m,n \in \mathbb{Z} such that m|n and \mathsf{gcd}(a,m) = 1. Then there exists \overline{a} \in \mathbb{Z} such that \overline{a} \equiv a mod m and \mathsf{gcd}(\overline{a},n) = 1. Proof: Let d = \mathsf{gcd}(a,n) and write n = mq. Now \mathsf{gcd}(d,m) = 1, so that d|q; we also write a = a^\prime d and q = q^\prime d. Let t be the product of all prime divisors of q^\prime which do not divide d. Finally, let \overline{a} = a + tm. Suppose p is a prime divisor of n. There are three cases:

    1. If p|m, then p \not| a since a and m are relatively prime. Thus p does not divide a + tm = \overline{a}.
    2. If p \not|m and p | q^\prime, we have two cases.
      1. If p|d, then p \not| t by definition. Thus p \not| tm. Also, since p|d, we have p|a. Thus p \not| a + tm \overline{a}.
      2. If p \not| d, then p|t and p \not| a. Thus p \not| a + tm = \overline{a}.
    3. If p \not| m,q^\prime, then since n = mq^\prime d, p | d. Now p | a and p \not| t, so p \not| a + tm = \overline{a}.

    Since no prime divisor of n divides \overline{a}, \mathsf{gcd}(\overline{a},n) = 1. \square

    Now to the main result.

    Suppose K \cong Z_n is finite. Now \mathsf{im}\ \varphi_1 is cyclic of order m where m|n, by Lagrange. Since x generates K, \varphi_1(x) generates \varphi_1[K]. Since conjugation by \sigma is an isomorphism \varphi_1[K] \rightarrow \varphi_2[K], \varphi_2(x)^a generates \varphi_2[K]. Thus \mathsf{gcd}(a,m) = 1. By the Lemma, there exists \overline{a} such that \overline{a} \equiv a mod m and \mathsf{gcd}(\overline{a},n) = 1. Moreover, there exists b such that \overline{a}b \equiv 1 mod n.

    Define \chi : H \rtimes_{\varphi_2} K \rightarrow H \rtimes_{\varphi_1} K by \chi((h,k)) = (\sigma^{-1}(h), k^b). This map is clearly a two sided inverse of \psi; hence H \rtimes_{\varphi_2} K \cong H \rtimes_{\varphi_1} K.

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  • Rahul  On December 27, 2010 at 11:18 pm

    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?

    • nbloomf  On December 28, 2010 at 8:44 am

      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 H = Z_8, since then \mathsf{Aut}(H) \cong Z_2 \times Z_2 has three nonconjugate subgroups of order 2 that we can use for K. 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.

      • Rahul  On January 9, 2011 at 12:59 am

        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?

        • nbloomf  On January 9, 2011 at 2:04 am

          Perhaps this works.

          Let H,K be groups and \varphi,\psi : K \rightarrow \mathsf{Aut}(H) group homomorphisms. Suppose \Phi : H \rtimes_\varphi K \rightarrow H \rtimes_\psi K is a group isomorphism. We want to show that \varphi[K] and \psi[K] are conjugate in \mathsf{Aut}(H).

          As a set function, \Phi(h,k) = (\Phi_1(h), \Phi_2(k)) for some bijections \Phi_1 : H \rightarrow H and \Phi_2 : K \rightarrow K. I claim that in fact \Phi_1 and \Phi_2 are isomorphisms. To see this, note that for all (h_1,k_1), (h_2,k_2) \in H \rtimes_\varphi K, we have (\Phi_1(h_1\varphi(k_1)(h_2)), \Phi_2(k_1k_2)) = \Phi(h_1 \varphi(k_1)(h_2), k_1k_2) = \Phi((h_1,k_1)(h_2,k_2)) = \Phi(h_1,k_1) \Phi(h_2,k_2) = (\Phi_1(h_1),\Phi_2(k_1))(\Phi_1(h_2),\Phi_2(k_2)) = (\Phi_1(h_1) \psi(\Phi_2(k_1))(\Phi_1(h_2)), \Phi_2(k_1)\Phi_2(k_2)). Comparing entries, we see that \Phi_2(k_1k_2) = \Phi_2(k_1)\Phi_2(k_2) for all k_1,k_2 \in K. So \Phi_2 is an automorphism of K. Now fix k_1 = 1; then \varphi(k_1) = \mathsf{id} and \psi(\Phi_2(k_1)) = \mathsf{id}. Thus \Phi_1(h_1h_2) = \Phi_1(h_1)\Phi_1(h_2), so that \Phi_1 is an automorphism of H.

          Next, we claim that \Phi_1 conjugates \varphi[K] into \psi[K]. To see this, recall the equality of first coordinates above and let h_1 = 1 and h_2 = h. Then \Phi_1(h_1 \varphi(k)(h)) = \psi(\varphi_2(k))(\Phi_1(h)) for all k and h; equivalently, we have \Phi_1 \circ \varphi(k) = \psi(\Phi_2(k)) \circ \Phi_1 for all k, and since \Phi_1 is an isomorphism, \Phi_1 \circ \varphi(k) \circ \Phi_1^{-1} = \psi(\Phi_2(k)).

          In particular, \Phi_1 \circ \varphi[K] \circ \Phi_1^{-1} \subseteq \psi[K]. Likewise, since \Phi_2 is an automorphism of K, we have \Phi_1^{-1} \circ \psi[K] \circ \Phi_1 \subseteq \varphi[K], and in fact \psi[K] = \Phi_1 \circ \varphi[K] \circ \Phi_1^{-1}.

          • nbloomf  On January 9, 2011 at 9:32 pm

            Eh… never mind. I’m not so sure anymore about the line \Phi(h,k) = (\Phi_1(h), \Phi_2(k)). Perhaps this can be fixed, though.

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