# Proving $G_{pq}$ is a subgroup.

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.

I want to prove that $$\begin{equation*} G_{pq}:=\left\{ \begin{bmatrix} a & b \\ 0 & 1 \end{bmatrix} \in GL_2(\mathbb{Z}_q) \mid a^{p}=1 \right\} \end{equation*}$$ is a nonabelian subgroup of $$GL_2(\mathbb{Z}_q)$$.

$$G_{pq}$$ is obviously nonabelian and associativity holds. The identity is $$\begin{bmatrix}1&0\\0&1\end{bmatrix}$$. I want to check if inverse of $$\begin{bmatrix}a&b\\0&1\end{bmatrix}$$ is $$\begin{bmatrix}a^{-1}&-a^{-1}b\\0&1\end{bmatrix}$$. Am I right? But I am not using condition $$q\equiv 1 \pmod{p}$$, so I think I am missing something here.

• What is $b$ in the set definition? – ajotatxe Oct 9 at 17:28
• @ajotatxe There was no definition of $b$ in this question other than $b\in\mathbb{Z}_q$. – aloevera Oct 9 at 17:32
• You need to check that it is closed under multiplication. That is probably where the condition will come up. – Matt Samuel Oct 9 at 17:45
• It's also not obvious that it is nonabelian. If $a$ were always equal to $1$, it would be abelian. You need to produce an example of two elements that don't commute. – Matt Samuel Oct 9 at 17:49
• If $q\not\equiv 1\pmod{p}$ then $a^p=1$ implies $a=1$ (and then... see above). – metamorphy Oct 9 at 18:24