Show that the semi-group of matrices $\begin{bmatrix}x&0\\ y&1\end{bmatrix}$ with $x>0$ and $y>0$, is simple Let $S=\left\{\begin{bmatrix}x&0\\ y&1\end{bmatrix}: x,y\in (0,\infty)\right\}$ be a semigroup under matrix multiplication. How to prove $S$ is simple but neither left nor right simple?
the subsets of $S$ defined by the conditions $y>1$ and $y>2x$  are respectively left and right ideals. But I got stuck in proving $S$ is simple. If $\emptyset\neq T\subset S$ is an ideal we should show $T=S$. for arbitrary element $\begin{bmatrix}a&0\\ b&1\end{bmatrix}\in S$ we must prove it's an element of $T$ by using $TS\subset T$ and $ST\subset T$. I tried many matrices with no luck.
Definitions:
$\emptyset\neq T\subset S$ is said to be


*

*left ideal if $ST\subset T$

*right ideal if $TS\subset T$ and

*an ideal if it is both left and right ideal


$S$ is simple if it has no proper ideal.
 A: Following conditions are equivalent:
$\textbf{1})$ $S$ is simple semigroup. 
$\textbf{2})$ $SaS=S$ for all $a\in S$. 
$\textbf{3})$ for all $a,b\in S$ there exist $s,t\in S$ such that $sat=b$
$\textbf{4})$ $J=S\times S$ where  $J$ is a green  relation.
$\textbf{Note:}$ Green relation means $J$ is equivalance relation such that $aJb\iff S^1aS^1=S^1bS^1$
Above conditions can be proved easily.
I will show  condition $\textbf{2)}$ i.e., $SxS=S$ for every $x\in S$ and note that every element of $x\in S$ invertible because determinant of $x$ greater than $1$.
Let $x=\begin{bmatrix}a&0\\ b&1\end{bmatrix}\in S$ and $g,h>0$. 
If we take $d=1, e=\frac{ha}{(a+b)^2},c=g\frac{(a+b)^2}{ha}\frac{1}{a},f=\frac{hb}{a+b}$ then we obtain
$\begin{bmatrix}c&0\\ d&1\end{bmatrix}\begin{bmatrix}a&0\\ b&1\end{bmatrix}\begin{bmatrix}e&0\\ f&1\end{bmatrix}=\begin{bmatrix}cae&0\\ dae+be+f&1\end{bmatrix}=\begin{bmatrix}g\frac{(a+b)^2}{ha}\frac{1}{a}a\frac{ha}{(a+b)^2}&0\\ \frac{ha(a+b)}{(a+b)^2}+\frac{hb}{a+b}&1\end{bmatrix}=\begin{bmatrix}g&0\\ h&1\end{bmatrix}$.
Thus for any $x\in S$, we have $SxS=S$, so $S$ is simple.
