Is $G = \left\{ \begin{bmatrix}a & -b \\ b & a\end{bmatrix} : a,b \in \Bbb R \right\}$ under matrix multiplication a monoid, semigroup, or neither?

Question

Determine whether $( G , \ast )$ is a monoid or semigroup or neither, where $$G = \left\{ \begin{bmatrix}a & -b \\ b & a\end{bmatrix} : a,b \in \Bbb R \right\}$$ and $\ast$ is product of matrices

i tired solve it :

by ${ x*x^{-1} }$ = I = ${ \begin{bmatrix}{1} && {0} \\ {0} && {1}\end{bmatrix} }$ but i get finally i =${ \begin{bmatrix}{aa} && {0} \\ {0} && {bb}\end{bmatrix} }$ is my steps is Right ? is these Group according to ${ x*x^{-1} = e }$ ?

Answer 1

It is a monoid (and hence also a semigroup).

First, we have to show that $I_2 \in G$. Indeed, setting $a=1$ and $b=0$ gives the identity matrix.

Second, we have to show that $G$ is closed under multiplication. Suppose that $\begin{bmatrix} a&-b \\ b&a \end{bmatrix} \in G$ and $\begin{bmatrix} c&-d \\ d&c \end{bmatrix} \in G$. Then, we have

$$\begin{bmatrix} a&-b \\ b&a \end{bmatrix}\begin{bmatrix} c&-d \\ d&c \end{bmatrix}=\begin{bmatrix} ac-bd&-(ad+bc) \\ ad+bc&ac-bd \end{bmatrix} \in G$$, so $G$ is closed under multiplication.

Finally, multiplication of matrices is associative in general, so it is in particular associative for matrices in $G$.

Alternatively, we could instead use the fact that $G$ is isomorphic to the monoid of complex numbers under multiplication via the map $\begin{bmatrix} a&-b \\ b&a \end{bmatrix} \mapsto a+bi$, and the fact that multiplication of complex numbers is associative.

Answer 2

It cannot be a semigroup without being a monoid since $$\begin{pmatrix} 1 & 0\\ 0 & 1\end{pmatrix}\in G$$ by setting $a=1$ and $b=0$; that is, there is an identity element in $G$ with respect to $\ast$.

What's left is to check whether $(G, \ast)$ is closed and associative.