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Let $p$ and $q$ be prime numbers such that $q \equiv 1(\text{mod} \ p)$. It is known that a group of order $p \cdot q$ is either isomorphic to $$\mathbb{Z}_{p} \times \mathbb{Z}_{q} \cong \mathbb{Z}_{pq}$$ or $$\mathbb{Z}_{q} \rtimes_{\phi} \mathbb{Z}_{p}$$ where $$\phi\colon \mathbb{Z}_{p} \to \text{Aut}(\mathbb{Z}_{q})$$ is a non-trivial morphism. We remark that $\mathbb{Z}_{q} \rtimes_{\phi} \mathbb{Z}_{p}$ is non-abelian.

So, it is natural to ask: given a non-abelian group $G$ of order $p \cdot q$ such that $q \equiv 1(\text{mod} \ p)$, what is the center of $G$?

Using the class equation, it is easily seen that a group of prime power order has a non-trivial center. Can the class equation be used to show that a group of the form $\mathbb{Z}_{q} \rtimes_{\phi} \mathbb{Z}_{p}$ has a non-trivial center?

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    $\begingroup$ The center is trivial. Otherwise, the center would have prime index, and the quotient $G/Z(G)$ would be cyclic, which is impossible in a non-abelian group. $\endgroup$ – verret Oct 28 '16 at 5:43
  • $\begingroup$ @verret Yes, that is correct. Thank you for your response. $\endgroup$ – John M. Campbell Oct 28 '16 at 19:46

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