Isomorphisms among semidirect products by a cyclic group
Let \(K\) be a cyclic group, \(H\) an arbitrary group, and \(\alpha\) and \(\beta\) homomorphisms \(K \rightarrow \mathsf{Aut}(H)\). If \(K\) is infinite, suppose further that \(\alpha\) and \(\beta\) are injective.
Show that if \(\mathsf{im}\ \alpha\) and \(\mathsf{im}\ \beta\) are conjugate subgroups in \(\mathsf{Aut}(H)\), then \[H \rtimes_{\alpha} K \cong H \rtimes_{\beta} K.\]
Let \(x\) be a generator of \(K\) and say that \(\sigma\) conjugates \(\mathsf{im}\ \alpha\) into \(\mathsf{im}\ \beta\); that is, that \(\sigma\left(\mathsf{im}\ \alpha\right)\sigma^{-1} = \mathsf{im}\ \beta\).
In particular, \(\sigma \alpha(x) \sigma^{-1} \in \mathsf{im}\ \beta\), so that \(\sigma \alpha(x) \sigma^{-1} = \beta(x^a)\) for some (not necessarily unique!) \(x^a \in K\). Now if \(x^b\) is any element of \(K\), we have \[\sigma \alpha(x^b) \sigma^{-1} = \sigma \alpha(x)^b \sigma^{-1} = \left(\sigma \alpha(x) \sigma^{-1}\right)^b = \beta(x)^{ab} = \beta(x^b)^a.\] In particular, for all \(k \in K\) we have \(\sigma \alpha(k) = \beta(k)^a \sigma\).
Now define a mapping \(\psi : H \rtimes_{\alpha} K \rightarrow H \rtimes_{\beta} K\) by \(\psi(h,k) = (\sigma(h), k^a)\). We claim that \(\psi\) is a group homomorphism.
To see this, let \((h_1,k_1), (h_2,k_2) \in H \rtimes_{\varphi\_1} K\). We have \[\begin{array}{rcl} \psi\left((h_1,k_1) \cdot (h_2,k_2)\right) & = & \psi\left(h_1 \alpha(k_1)(h_2), k_1k_2\right) \\ & = & \left(\sigma(h_1 \alpha(k_1)(h_2)), (k_1k_2)^a\right) \\ & = & \left(\sigma(h_1) \sigma(\alpha(k_1)(h_2)), k_1^a k_2^a\right) \\ & = & \left(\sigma(h_1) (\sigma \circ \alpha(k_1))(h_2), k_1^a k_2^a\right) \\ & = & \left(\sigma(h_1) (\beta(k_1)^a \sigma)(h_2), k_1^a k_2^a\right) \\ & = & \left(\sigma(h_1) \beta(k_1^a)(\sigma(h_2)), k_1^a k_2^a\right) \\ & = & \left(\sigma(h_1), k_1^a)(\sigma(h_2), k_ 2^a\right) \\ & = & \psi(h_1,k_1) \cdot \psi(h_2,k_2), \\ \end{array}\] and thus \(\psi\) is a homomorphism.
We now show that \(\psi\) is bijective.
First suppose \(K\) is infinite, so that (by hypothesis) both \(\alpha\) and \(\beta\) are injective. Just as \(\sigma \alpha(k) = \beta(k)^a \sigma\) for all \(k \in K\), there is an integer \(b\) such that \(\beta(k) \sigma = \sigma \alpha(k)^b\) for all \(k \in K\). Combining these results we see that \(\beta(k) = \beta(k^{ab})\). Since \(\beta\) is injective we have \(k^{1-ab} = 1\), and since \(K\) is infinite and \(k\) arbitrary, we have \(ab = 1\). Thus \(a = b \in \{1,-1\}\); in either case we have \(a^2 = 1\).
Define a mapping \(\chi : H \rtimes_{\beta} K \rightarrow H \rtimes_{\alpha} K\) by \(\chi(h,k) = \left(\sigma^{-1}(h), k^a\right)\). 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 \(H \rtimes_{\beta} K \cong H \rtimes_{\alpha} K\).
Before proceeding to the finite case we prove the following lemma, due to Luís Finotti.
Let \(a\), \(m\), and \(n\) be integers such that \(m \mid n\) and \(\mathsf{gcd}(a,m) = 1\). Then there is an integer \(\overline{a}\) such that \(\overline{a} \equiv a \pmod{m}\) and \(\mathsf{gcd}(\overline{a},n) = 1\).
Let \(d = \mathsf{gcd}(a,n)\) and write \(n = mq\). Note that \(\mathsf{gcd}(d,m) = 1\) since \(d\) divides \(a\), so that \(d \mid q\) by Euclid’s Lemma. Write \(a = a^\prime d\) and \(q = q^\prime d\), and let \(t\) be the (possibly empty) product of all prime divisors of \(q^\prime\) which do not divide \(d\). Finally, let \(\overline{a} = a + tm\). Certainly \(\overline{a} \equiv a \pmod{m}\).
Suppose \(p\) is a prime divisor of \(n\). We claim that \(p\) does not divide \(\overline{a}\); to show this we consider three possibilities.
- If \(p \mid m\), then \(p \not\mid a\) since \(a\) and \(m\) are relatively prime. Thus \(p\) cannot divide \(a + tm = \overline{a}\).
- If \(p \not\mid m\) and \(p \mid q^\prime\), we have two cases.
- If \(p \mid d\), then \(p \mid a\). We also have \(p \not\mid t\) by construction, so that \(p \not\mid tm\). Thus \(p\) cannot divide \(a + tm = \overline{a}\).
- If \(p \not\mid d\), then \(p \mid t\) by construction. Note also that \(p \not\mid a\), since otherwise \(p\) divides both \(a\) and \(n\) and so must divide \(d\). Thus \(p\) cannot divide \(a + tm = \overline{a}\).
- If \(p \not\mid m\) and \(p \not\mid q^\prime\), then since \(n = m q^\prime d\) we have that \(p \mid d\) and thus \(p \mid a\). Now \(p \not\mid t\) by construction, so that \(p \not\mid tm\). Thus \(p\) cannot divide \(a + tm = \overline{a}\).
Since no prime divisor of \(n\) divides \(\overline{a}\) we have \(\mathsf{gcd}(\overline{a},n) = 1\).
Now to the main result.
Suppose \(K \cong Z_n\) is finite. Note that \(\mathsf{im}\ \alpha\) is cyclic of order \(m\) for some \(m \mid n\) by Lagrange’s Theorem. Since \(x\) generates \(K\), \(\alpha(x)\) generates \(\mathsf{im}\ \alpha\), and since conjugation by \(\sigma\) is an isomorphism \(\mathsf{im}\ \alpha \rightarrow \mathsf{im}\ \beta\), we have that \(\sigma\alpha(x)\sigma^{-1} = \beta(x)^a\) generates \(\mathsf{im}\ \beta\). Thus \(\mathsf{gcd}(a,m) = 1\). By the lemma, there exists \(\overline{a}\) such that \(\overline{a} \equiv a \pmod{m}\) and \(\mathsf{gcd}(\overline{a},n) = 1\). Moreover, there exists \(b\) such that \(\overline{a}b \equiv 1 \pmod{n}\).
Define a map \(\chi : H \rtimes_{\beta} K \rightarrow H \rtimes_{\alpha} K\) by \(\chi(h,k) = (\sigma^{-1}(h), k^b)\). It is straightforward to check that this map is a two-sided inverse of \(\psi\); hence \(H \rtimes_{\beta} K \cong H \rtimes_{\alpha} K\).