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.

  • $\begingroup$ What is $b$ in the set definition? $\endgroup$ – ajotatxe Oct 9 at 17:28
  • $\begingroup$ @ajotatxe There was no definition of $b$ in this question other than $b\in\mathbb{Z}_q$. $\endgroup$ – aloevera Oct 9 at 17:32
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    $\begingroup$ You need to check that it is closed under multiplication. That is probably where the condition will come up. $\endgroup$ – Matt Samuel Oct 9 at 17:45
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    $\begingroup$ 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. $\endgroup$ – Matt Samuel Oct 9 at 17:49
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    $\begingroup$ If $q\not\equiv 1\pmod{p}$ then $a^p=1$ implies $a=1$ (and then... see above). $\endgroup$ – metamorphy Oct 9 at 18:24

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