I am studying group theory and encountered this problem.

If $q=1 \pmod{p}$, then every group of order $pq$ is either cyclic or isomorphic to $G_{p,q}$.

Here $G_{p,q}$ is defined as follows: let $p, q$ be distinct primes with $q\equiv 1 \pmod{p}$. Let $\mathbb{Z}_q^*$ be the group of nonzero elements of $\mathbb{Z}_q$ under multiplication.

\begin{equation*} G_{pq}:=\left\{ \begin{bmatrix} a & b \\ 0 & 1 \end{bmatrix} \in GL_2(\mathbb{Z}_q) \mid a^{p}=1 \right\} \end{equation*}

I don't know where to start. Can anyone help me solving this problem?

I edited the problem.

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    $\begingroup$ What is $G_{p,q}$? $\endgroup$ – Randall Oct 8 at 13:58
  • $\begingroup$ @Randall It's a cyclic group $Z_{pq}$. Hungerford used this notation in his algebra book. $\endgroup$ – aloevera Oct 8 at 14:03
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    $\begingroup$ Then your conclusion is redundant: $G$ is either cyclic or it's cyclic. That can't possibly be what he means. $\endgroup$ – Randall Oct 8 at 14:11
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    $\begingroup$ My guess is that $G_{p,q}$ is the non-abelian group with generators $x, y$ satisfying order of $x$ is $p$, order of $y$ is $q$, and $yx=x^ty$ for some correctly chosen $t$. $\endgroup$ – Randall Oct 8 at 14:18
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    $\begingroup$ When $p=2$ we get $q$ must be odd, and this boils down to the proof that a non-abelian group of order $2q$ must be dihedral. Your problem is a generalization of this simpler fact. You could start there to gain an understanding. $\endgroup$ – Randall Oct 8 at 14:36

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