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

2018-02-17 06:47:32

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 $$\begin{bmatrix}1 & n \\ 0 & 1\end{bmatrix}\begin{bmatrix}1 & -n \\ 0 & 1\end{bmatrix}$$

    2018-02-17 07:42:59
  • This looks isomorphic to the group $(\mathbb Z,+)$ to me.

    2018-02-17 08:51:59