Prove that $G$ is a finitely generated group under multiplication. 
Consider the set of matrices $$G = \left \{\begin{pmatrix} s & b \\ 0 & 1 \end{pmatrix}\ \bigg |\ b \in \Bbb Z,\ s \in \left \{-1,+1 \right \} \right \}.$$ Then show that $G$ is a finitely generated group under multiplication.

It is easy to show that $G$ is a group under multiplication. What I think is that $$G = \left \langle \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} -1 & 1 \\ 0 & 1 \end{pmatrix} \right \rangle.$$
The reason behind my claim is that the elements of $G$ are of the form $\begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix}$ or $\begin{pmatrix} -1 & n \\ 0 & 1 \end{pmatrix},$ for some $n \in \Bbb Z.$ Now for any $n \in \Bbb Z$ we have $$\begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix} = {\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}}^n.$$ Also if $n$ is an odd integer then $$\begin{pmatrix} -1 & n \\ 0 & 1 \end{pmatrix} = {\begin{pmatrix} -1 & 1 \\ 0 & 1 \end{pmatrix}}^n;$$ and if $n$ is an even integer then $$\begin{pmatrix} -1 & n \\ 0 & 1 \end{pmatrix} =  {\begin{pmatrix} -1 & 1 \\ 0 & 1 \end{pmatrix}}^{n+1} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} ;$$ which proves the claim.
Is my above reasoning correct at all? Can anybody please verify it?
Thanks in advance.

Well, you need to consider what happens when the generators multiply:
$\begin{pmatrix} -1 & 1 \\ 0 & 1\end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & 1\end{pmatrix}$ and
$\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix} \begin{pmatrix} -1 & 1 \\ 0 & 1\end{pmatrix} = \begin{pmatrix} -1 & 2 \\ 0 & 1\end{pmatrix}$.
