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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?
Closure under matrix multiplication:

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 $$\begin{bmatrix}1 & n \\ 0 & 1\end{bmatrix}\begin{bmatrix}1 & n \\ 0 & 1\end{bmatrix}$$
20180217 07:42:59 
This looks isomorphic to the group $(\mathbb Z,+)$ to me.
20180217 08:51:59