I am trying to show that $G$ is a subgroup of $GL_2(\mathbb{Z}_5)$ where $G$ is the set of all matrices in $GL_2(\mathbb{Z}_5)$ (the set of all invertible 2×2 matrices over $\mathbb{Z}_5$) of the form $\begin{pmatrix}
m & b\\
0 & 1
\end{pmatrix}$.
Starting with the inverse element I get:
$\begin{pmatrix}
m & b\\
0 & 1
\end{pmatrix}^{-1} = \begin{pmatrix} \frac{1}{m} & \frac{-b}{m} \\ 0 & 1 \end{pmatrix}$.
Clearly the bottom row of this inverse matrix is of the form we desire, but I'm confused about the top row and the $\mathbb{Z}_{5}$ condition. Obviously when $m=1$ and $b=0$ this inverse belongs to the subgroup, but does this need to be satisfied for all possible values of $m$ and $b$ in $\mathbb{Z}_5$ or is its existence with these values enough to satisfy the inverse condition of a subgroup?
Apologies if this is a dumb question and I've missed something obvious.
Best Answer
$m,n\in \mathbb{Z}_5\setminus \{0\} \Longrightarrow\begin{pmatrix} m & b\\ 0 & 1 \end{pmatrix} \cdot \begin{pmatrix} n & c \\ 0 & 1 \end{pmatrix}=\begin{pmatrix} mn & mc+b \\ 0 & 1 \end{pmatrix}\in G$, because $mn\in \mathbb{Z}_5\setminus \{0\}$.