# Isomorphisms among semidirect products by a cyclic group

Posted on 2010-06-25 by nbloomf

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.

1. If $$p \mid m$$, then $$p \not\mid a$$ since $$a$$ and $$m$$ are relatively prime. Thus $$p$$ cannot divide $$a + tm = \overline{a}$$.
2. If $$p \not\mid m$$ and $$p \mid q^\prime$$, we have two cases.
1. 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 $$a + tm = \overline{a}$$.
2. 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 $$a + tm = \overline{a}$$.
3. 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 $$H \rtimes_{\beta} K \cong H \rtimes_{\alpha} K$$.