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


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 $G$.


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:


Thanks in advance

  • 1
    $\begingroup$ The integers modulo $5$ form a field: each nonzero element has a multiplicative inverse. $\endgroup$ Apr 18, 2020 at 23:34
  • $\begingroup$ Hi @AnginaSeng thanks for the response. I am yet to cover fields in terms of number and group theory so I am not too sure what you mean by that. Also, do you know why is my argument wrong? $\endgroup$ Apr 18, 2020 at 23:37
  • 1
    $\begingroup$ Can you give an example of an element of $G$ lacking an inverse? $\endgroup$ Apr 18, 2020 at 23:40
  • 2
    $\begingroup$ @Kanye West: The exam problem is not incorrect. Note that the exam has $a\ne 0$ and $c\ne 0$, so you have a typo in your statement of the problem. Also, all matrix entries can be expressed as elements of the set $\{0,1,2,3,4\}$ with all operations done mod $5$. $\endgroup$
    – quasi
    Apr 18, 2020 at 23:42
  • 1
    $\begingroup$ Also, for the inverse you should not divide the way you do; you should use the multiplicative inverse operation inherited from the fact that integers mod 5 form a field as noted by Angina. $\endgroup$
    – E-A
    Apr 18, 2020 at 23:52

1 Answer 1


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).

  • $\begingroup$ Hi thanks for the answer. I have a few questions. Could you elaborate on why $\frac{1}{ac}$ is well defined as I am inexperienced in combining discrete and linear algebra? Also, it is not given that $a,c,d,f \neq 0$, is it implied by the question saying they are intigers thus discounting 0? Thanks in advance. $\endgroup$ Apr 18, 2020 at 23:56
  • $\begingroup$ From what quasi wrote the exam has $a\not=0$ and $c\not=0$ so $\frac{1}{ab}$ is well-defined since it is not 0. Similarly as $a\not=0$ and $c\not=0$ by assumption if we introduce a second matrix from G as you've done for closure, then $d,f\not=0$ as well. $\endgroup$
    – Alessio K
    Apr 19, 2020 at 0:00
  • $\begingroup$ You are right! Just a quick question. If I have for example an element of the matrix being $\frac{1}{3}$ would that translate to $1 \mod{3}$ which translates to $1 \mod {5}$ or $4 \mod{5}$ and moreover if that logic is correct doesn't that imply that the inverse is not unique? Thanks in advance. $\endgroup$ Apr 19, 2020 at 0:03
  • $\begingroup$ In this case if you wanted to find the inverse of 3 mod 5, you would write $3^{-1}=2 mod 5$ since $3(2)=6=1mod5$. $\endgroup$
    – Alessio K
    Apr 19, 2020 at 0:13
  • 1
    $\begingroup$ $\frac{1}{3}$ does not mean 1 mod 3. The question says the values are mod 5, that doesn't change just because you have a fraction. $\frac{1}{3}$ mod 5 means the inverse of 3 mod 5. $\endgroup$
    – Ted
    Apr 19, 2020 at 0:53

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