Let $p$ and $q$ be prime numbers such that $q$ divides $p-1$.

It is well-know that there is a monomorphism $\varphi: \mathbb Z/q \mathbb Z \to Aut(\mathbb Z/p \mathbb Z)$.

Define homomorphisms $\varsigma: (\mathbb Z/q \mathbb Z)^2 \to \mathbb Z/ q \mathbb Z$ where $(a,b) \mapsto a-b$ and $\vartheta: (\mathbb Z/q \mathbb Z)^2 \to Aut(\mathbb Z/p \mathbb Z)$ via $\vartheta = \varphi \circ \varsigma$.

Note that composition of maps is evaluated from right to left.


If we consider the semi-direct products $G := (\mathbb Z/p \mathbb Z \rtimes_\varphi \mathbb Z/q \mathbb Z) \times \mathbb Z/q \mathbb Z$ and $H := \mathbb Z/p \mathbb Z \rtimes_\vartheta (\mathbb Z/q \mathbb Z)^2$, are these groups isomorphic?


My intuition says: No, $G$ and $H$ are not isomorphic. But I am unsure how to prove this hypothesis. I tried to evaluate the centers $Z(G)$ and $Z(H)$ of $G$ and $H$, respectively, which gave me $Z(G) = \{(0,0)\} \times \mathbb Z/q \mathbb Z$ and $\{(0,r,r): r \in \mathbb Z/q \mathbb Z\} \subseteq Z(H)$.

Makes this line of attack sense? Or is the required argument quite obvious?

Thank you very much for your insights!


I stumbled upon this question while reading a collection of problems about group theory which interested me as a layperson.


I think that your groups are isomorphic. You can prove it in two ways. The first one is to show an explicit isomorphism, which is not hard to do, but kind of boring. The second way is to show that $H$ "has the same structure as $G$".

Namely you have to find two commuting normal subgroups of H, whose intersection is trivial, one isomorphic to $\mathbb{Z}/p\mathbb{Z} \rtimes_\varphi \mathbb{Z}/q\mathbb{Z}$ and the other isomorphic to $\mathbb{Z}/q\mathbb{Z}$. It's not hard to find the former. As for the latter, you have already found it!

  • $\begingroup$ Thank you! If your are right, I was looking in the wrong direction. I will try to your argument in the following days. Until then +1 from me. $\endgroup$ – Moritz Nov 18 '18 at 15:04

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