I have been doing some STEP papers and I think I found a mistake on one of the papers. The paper is from 1991 (STEP 2) and the question (Q9) goes as follows:
Let $G$ be the set of all matrices of the form $ \ \left[ {\begin{array}{cc} a & b \\ 0 & c \\ \end{array} } \right] \ $ where a,b and c are integers modulo $5$, $a\neq0\neq b$. Show that $G$ forms a group under matrix multiplication (which may be assumed to be associative).
I've proven that the identity exists and is unique, however, I can't prove that the inverses are going to exist within G and ,moreover, that G is closed. Here are my arguments
Inverses
Suppose there exists $A$ such that $A \times \ \left[ {\begin{array}{cc} a & b \\ 0 & c \\ \end{array} } \right] \ $ $= \ \left[ {\begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} } \right] \ $ . It follows that $A= \ \left[ {\begin{array}{cc} a & b \\ 0 & c \\ \end{array} } \right]^{-1} \ $ $=\frac{1}{ac} \ \left[ {\begin{array}{cc} c & -b \\ 0 & a \\ \end{array} } \right] \ $. The issues that arise are that $\frac{c}{ac}$ is not nesseraly an intiger for intiger c and similarly $\frac{a}{ac}$ is also not necceserally an intiger. Thus for $a\neq 1\neq c$ the inverses are not a member of $G$.
Closure
Let $ \ \left[ {\begin{array}{cc} a & b \\ 0 & c \\ \end{array} } \right] \ $ and $ \ \left[ {\begin{array}{cc} d & e \\ 0 & f \\ \end{array} } \right] \ $ be members of $G$. In doing matrix multiplication we get $ \ \left[ {\begin{array}{cc} a & b \\ 0 & c \\ \end{array} } \right] \ \times \ \left[ {\begin{array}{cc} d & e \\ 0 & f \\ \end{array} } \right] \ = \ \left[ {\begin{array}{cc} ad & ae+bf \\ 0 & cf \\ \end{array} } \right] \ $
We see that if $d$ is a multiple of $5$ then $ad\equiv0 \mod{5}$. However, if the new matrix is formed $ad$ must obey $ad \neq 0$ which it obviously doesn't. Thus it is not closed.
Can someone have a look at my arguments and let me know what I have done wrong or if I am right ?
Here is the link to the full paper:
https://pastpapercache.blob.core.windows.net/ppppapers/step/STEP%20II%201991.pdf
Thanks in advance
When working with cogruences, we think of "division" as "multiplying by the multiplicative inverse". Since both $a\not=0$ and $c\not=0$, then $\frac{1}{ac}$ is well-defined and we have existence of inverses. Closure is also satisfied since $a,c,d,f\not=0$. (Note: G is not commutative).