Show that if matrices A and B are elements of G, then AB is also an element of G. 
Let $G$ be the set of $2 \times 2$ matrices of the form
  \begin{pmatrix} a & b \\ 0 & c\end{pmatrix}
  such that $ac$ is not zero. Show that if matrices $A$ and $B$ are elements of $G$, then $AB$ is also an element of $G$.

Do I just need to show that $AB$ has a non-zero determinant? 
 A: Let $$A = \begin{bmatrix} a & b \\ 0 & c\\ \end{bmatrix}, \;\;
B = \begin{bmatrix} e & f \\ 0 &g \\ \end{bmatrix}$$
where $ac\neq 0,\;\;eg \neq 0$. So $A, B \in G$.
Simply compute $AB= P$ and what to you get? 
Use the definition of matrix multiplication, and the fact that $ac \neq 0$ and $eg\neq 0$, and check to see if the lower left entry of your product matrix $P$ is, in fact, $0$.
Showing that $\det (AB) = \det(P) \neq 0$ is not your task. In fact, the  $$\det \begin{bmatrix} m & 0\\n& q\\ \end{bmatrix} \neq 0$$ when $m, n, q$ are non-zero, but this matrix is NOT in $G$.  
You need to verify that for the entries $p_{ij}$ of $AB = P$:
$p_{11}p_{22} \neq 0.$
$p_{21} = 0$.  
Once you've done that, you can conclude $AB = P \in G$.
A: Proving that AB has a non-zero determinant is not enough, because not all 2x2 matrices with non-zero determinant are a element of G.
You need to prove another property of AB. This property is that it has the shape you stated.
This combined with a non-zero determinant guarantees that AB has the prescribed shape with ac not zero.
A: Another way to put it: entry $G_{21}$ is given by the dot product of vectors $(0 \space A_{22})$ and $(B_{11} \space 0)$. These are orthogonal, ie their dot product is zero, so that entry is always 0.
