Explicit example of non abelian group of order $pq$ where $p \mid q-1$

Question

I want to show groups from semi-direct product of order $pq$ where $p \mid q-1$ is non-abelian.

I know $\mathbb{Z}_p \rtimes _\phi \mathbb{Z}_q$ is a group for non-trivial group homomorphism $\phi:\mathbb{Z}_p \to \mathrm{Aut} (\mathbb{Z}_q)$, and they are all the same. Many texts claim this group is non-abelian, but how do we know it is non-abelian?

Consider $\mathbb{Z}_3 \rtimes _\phi \mathbb{Z}_7$ where $\mathbb{Z}_3=\{0,1,2\}$ defined by $\phi(1_{\mathbb{Z}_3})=2_{\mathrm{Aut}(\mathbb{Z}_7)}$. So $\phi(2)=4$ and $x_{\mathrm{Aut}(Z_7)}(y_{Z_7})=xy_{Z_7}$. I just plugged in random values and found $$ (3,1)\cdot_\rtimes(4,2)=(3\phi(1)(4),2)=(3\cdot4\cdot2,2)=(3,2) \\ (4,2)\cdot_\rtimes(3,1)=(4\phi(2)(3),2)=(4\cdot3\cdot4,2)=(6,2) $$

So $\mathbb{Z}_3 \rtimes _\phi \mathbb{Z}_7$ is non abelian. Is there a way to generalize this to show all groups generated by non trivial homomorphism to automorphism is non-abelian? Or am I missing out a more simple elementary argument?

(Note: I’m not exactly sure if my notation is correct, please correct if there’s a mistake)

Answer 1

Note that $S_3=\mathbb{Z}_3\rtimes\mathbb{Z}_2$ is an easy example of a non-abelian groups of order $pq$. Let $H=\langle a\rangle\cong\mathbb{Z}_3$ and $K=\langle b\rangle\cong\mathbb{Z}_2$. Then $\operatorname{Aut}(H)$ has one non-trivial automorphism $\pi: 1\mapsto 1, a\mapsto a^2, a^2\mapsto a$. Conjugation by $b$ induces this automorphism, so $b^{-1}ab=a^2$ and $b^{-1}a^2b=a$.

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To answer your question: If we have a semidirect product $G=K\rtimes_{\phi} H$ where $H$ and $K$ are cyclic then $G$ is abelian if and only if the automorphism $\phi$ of $H$ is trivial. To see this, remember that $k^{-1}hk=\phi(h)$ for some generator $k$ of $K$.

1. If $\phi$ is trivial then $\phi(h)=h$ for all $h\in H$. Therefore, $k^{-1}hk=h$ for all $h\in H, k\in K$, and so every element of $H$ commutes with every element of $K$. Hence, $G$ is abelian as required.
2. If $\phi$ is non-trivial then there exists some $h\in H$ such that $\phi(h)\neq h$. Therefore, $k^{-1}hk\neq h$, and so $h$ and $k$ do not commute. Hence, $G$ is not abelian as required.