Let $n = p_1 p_2 \cdots p_k$ be the product of pairwise distinct odd primes. Let $X$ be the set of element in $\operatorname{Aut}(\mathbb Z/n\mathbb Z)$ of order $1$ or $2$. For each $\psi\in X$, the mapping $$\varphi_\psi : \mathbb Z/2\mathbb Z \to \operatorname{Aut}(\mathbb Z/n\mathbb Z),\quad x\mapsto \begin{cases}\operatorname{id} & \text{if }x = 0 \\ \psi & \text{if }x = 1\end{cases}$$ is a homomorphism of groups. So each $\psi\in X$ defines a semidirect product $$G_\psi = \mathbb Z/n\mathbb Z\rtimes_{\varphi_\psi} \mathbb Z/2\mathbb Z.$$ Are these groups $G_\psi$ pairwise non-isomorphic?

This question came to my mind when giving this answer on the classification of groups of order $30$.

We have \begin{align*}\operatorname{Aut}(\mathbb Z/n\mathbb Z) & \cong (\mathbb Z/n\mathbb Z)^\times \\ & \cong (\mathbb Z/p_1\mathbb Z)^\times \times \ldots \times (\mathbb Z/p_k \mathbb Z)^\times \\ & \cong \mathbb Z/(p_1 - 1)\mathbb Z\times \cdots \times \mathbb Z/(p_k - 1)\mathbb Z.\end{align*}

In the latter representation, the set $X$ is given by $$X = \left\{(a_1,\ldots,a_k) \mathrel{}\middle|\mathrel{} a_i\in\left\{0, \frac{p_i - 1}{2}\right\}\right\}$$ So $\lvert X\rvert = 2^k$.

It is clear that $G_\psi$ is abelian (even cyclic) if and only if $\psi = \operatorname{id}$. So for $k = 1$, the answer is yes. In the case $n = 3\cdot 5$ (which leads to the classification of the groups of order 30) the answer is yes as well, and I vaguely conjecture that the answer should be yes in general. So probably we should look at some kind of an invariant which separates the groups $G_\psi$.


For any such $G$, and each $i$, the element $\frac{n}{p_i} \in \mathbb{Z}/n\mathbb{Z}$, together with the generator of $\mathbb{Z}/2\mathbb{Z}$, generates a subgroup $G_i$ of the form $\mathbb{Z}/p_i\mathbb{Z} \rtimes \mathbb{Z}/2\mathbb{Z}$.

I claim that $\varphi_\psi$ can be reconstructed once you know the isomorphism type of each $G_i$.

If you believe my claim, then this is all you need, because there are only two possible semidirect products $\mathbb{Z}/p_i\mathbb{Z} \rtimes \mathbb{Z}/2\mathbb{Z}$: an abelian one and a non-abelian one.

Here's a little push towards seeing why this claim is true: the isomorphism type of $G_i$ is determined precisely by $\varphi_\psi (1) (\frac{n}{p_i})$.

  • $\begingroup$ I understand your clever argument that the subgroups $G_p$ allow you to reconstruct $\varphi_\psi$. But why does this answer my question? I think we need some additional way to characterize the subgroups $G_p$ without relying on a representation of $G$ as a concrete semidirect product. More concretely: Assume that $\Phi : G_\psi \to G_{\psi'}$ is an isomorphism ($\psi \neq \psi'\in X$). By your argument it is clear that $\Phi((G_\psi)_p) \neq G_p$ for some prime factor $p$ or $n$. But why shouldn't it be possible that $\Phi((G_\psi)_p)$ is some "skew" subgroup of $G_{\psi'}$? $\endgroup$ – azimut Nov 16 '13 at 23:29
  • $\begingroup$ @azimut You're right, there's a choice here, namely the choice of an element of order $2$. But an element of $\mathbb{Z}/n\mathbb{Z}$ commutes with one element of order $2$ if and only if it commutes with them all. Maybe a simpler way to say all of this is the following: which primes divide $|Z(G)|$? $\endgroup$ – Slade Nov 16 '13 at 23:43

For each of the $2^k - 1$ divisors $d\neq 1$ of $n$, we define the group $$H_d = D_d \times \mathbb Z/(n/d)\mathbb Z.$$

The groups $H_d$ are non-abelian groups of order $2n$, and because of $$ D_d \times \mathbb Z/(n/d)\mathbb Z \\ \cong (\mathbb Z/d\mathbb Z \rtimes \mathbb Z/2\mathbb Z) \times \mathbb Z/(n/d)\mathbb Z\\ \cong (\mathbb Z/d\mathbb Z\times \mathbb Z/(n/d)\mathbb Z) \rtimes \mathbb Z/2\mathbb Z\\ \cong \mathbb Z/n\mathbb Z \rtimes \mathbb Z/2\mathbb Z $$ the groups are of the form considered in the question.

Since $d$ is odd, there number of elements of order $1$ or $2$ in $D_d$ is $d$. Furthermore, since $n/d$ is odd, there are no elements of order $2$ in $Z/(n/d)\mathbb Z$.

This shows that the number of elements of order $2$ in $H_d$ is $d$. So the groups $H_d$ are pairwise non-isomorphic. Adding the abelian group $\mathbb Z/2n\mathbb Z$, we see that there are at least $2^k$ isomorphism types of semidirect products of the form $\mathbb Z/n\mathbb Z \rtimes \mathbb Z/2\mathbb Z$.

Since there are only $2^k$ possibilities to define a homomorphism $\mathbb Z/2\mathbb Z \to \operatorname{Aut}(\mathbb Z/n\mathbb Z)$ (see the question), we get that the number of isomorphism types of semidirect products $\mathbb Z/n\mathbb Z \rtimes \mathbb Z/2\mathbb Z$ is exactly $2^k$, and those given in the question are pairwise non-isomorphic.


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