## 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 $H \rtimes_{\varphi_2} K \cong H \rtimes_{\varphi_1} K$.

• 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.