# Is the $2\times 2$ matrix a group?

Consider $$M:=\Big\{ \left[ {\begin{array}{cc} 1 & n \\ 0 & 1 \\ \end{array} } \right] :n \in \mathbb Z \Big\}$$

Is $M$ a group under the operation of matrix multiplication?

I know that we have to show that $M$ is closed under matrix multiplication, which is associative in $M$, and there is an identity element in M and each element in $M$ has an inverse. I think that because the determinant will be positive, it is closed under matrix multiplication.

Moreover, as matrix multiplication is associative in general, it holds true here. and: $\ e= \left[ {\begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} } \right]$ is a clear identity. Further: any element $\ q= \left[ {\begin{array}{cc} a & 0 \\ 0 & b \\ \end{array} } \right] \in M, \ e= \left[ {\begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} } \right]$ is an inverse(I think??) And therefore it forms a group. QED

Any Thoughts?

• The set of matrices of the form $\pmatrix{1&n\\0&1}$ for $n\in\Bbb Z$ is a group. You need to prove that the inverse of each such matrix is also in the set. – Angina Seng Feb 17 '18 at 3:40
• Just pick generic matrices $\pmatrix{1 & n \\ 0 & 1}$ and $\pmatrix{1 & m \\ 0 & 1}$, and compute their product -- I think you'll have a much better understanding of what you're up against. – pjs36 Feb 17 '18 at 3:42
• $e$ is not an inverse-- it's the identity element. – coffeemath Feb 17 '18 at 3:44

Closure under matrix multiplication:

$$\begin{bmatrix}1 & n \\ 0 & 1\end{bmatrix}\begin{bmatrix}1 & m \\ 0 & 1\end{bmatrix}= \begin{bmatrix}1 & m+n \\ 0 & 1\end{bmatrix}$$

Since $m+n\in \mathbb{Z}$, it is closed under matrix mulitplication. That is if $A \in M$ and $B \in M$, then $AB \in M$.

• It is irrelevant to the determinant.

Identity and associativity inherits from the normal matrix operations.

• matrix $q$ again is irrelevant, afterall, in general , $q \notin M$.

For matrix inverse, try to evaluate $$\begin{bmatrix}1 & n \\ 0 & 1\end{bmatrix}\begin{bmatrix}1 & -n \\ 0 & 1\end{bmatrix}$$

This looks isomorphic to the group $(\mathbb Z,+)$ to me.

Another way to prove that $M$ is a group is to notice that $\begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix}=\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^{n}$(you can check this by induction)

So $M=<\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}>$