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

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.

• What is $G_{p,q}$? – Randall Oct 8 at 13:58
• @Randall It's a cyclic group $Z_{pq}$. Hungerford used this notation in his algebra book. – aloevera Oct 8 at 14:03
• Then your conclusion is redundant: $G$ is either cyclic or it's cyclic. That can't possibly be what he means. – Randall Oct 8 at 14:11
• 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 $t$. – Randall Oct 8 at 14:18
• 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. – Randall Oct 8 at 14:36