<s>[INST] Semi group presentation $$ Another semi group question here, trying to get my head around the topic.
Consider the semi group $S=\left<a, b | a^{2} = b^{2} = 0, aba = a, bab = b\right>$
I need to prove that $S$ has order $5$.
The way I did it was to consider $a$ and $b$: we have to apply either $a$ or $b$ to these elements so $a.a=0$ and $a.b=ab$ $b.a=ba$ $b.b=0$ gives us three new elements for so far we have $\{a,b,ab,0,ba\}$.  Now consider $ab$: we have to apply either $a$ or $b$ to this; applying $a$ we get $aba$ which is $b$ so not a new element; applying $b$ we get $abb=0$ so not a new element. Finally consider $ba$ (as we know applying $a$ or $b$ to $0$ gives $0$): applying $a$ to $ba$ we get $0$; applying $b$ to $a$ we get $ab$ which gives us no new elements.
Thus we have at most $5$ elements; are they all distinct though! I mean I can see they are but how do we know that say $ab \neq ba$ or whatever? How can I show that all these elements are distinct?
Cheers guys
 [/INST] This is a presentation of the Brandt semigroup $B_2$, which can be described 
as the semigroup generated by the two following matrices
$$
 a = \begin{pmatrix} 0&1 \\ 0&0 \end{pmatrix}, \quad
 b = \begin{pmatrix} 0&0 \\ 1&0 \end{pmatrix}, 
$$
under the usual multiplication of matrices. Thus $B_2 = \{a, b, ab, ba, 0\}$ with
$$ 
 a = \begin{pmatrix} 0&1 \\ 0&0 \end{pmatrix}, \quad
 b = \begin{pmatrix} 0&0 \\ 1&0 \end{pmatrix}, \quad
 ab = \begin{pmatrix} 1&0 \\ 0&0 \end{pmatrix}, \quad
 ba = \begin{pmatrix} 0&0 \\ 0&1 \end{pmatrix}, \quad
 0 = \begin{pmatrix} 0&0 \\ 0&0 \end{pmatrix}
$$
The relations $a^2= 0$, $b^2 = 0$, $aba = a$ and $bab = b$ are trivially satisfied.
</s>