Brandt Semigroup : Consider set $[n] = \{ 1,2,\cdots ,n\}$ , then $B_n = [n] \times [n] \cup \{0\} $ is a semigroup under the binary operation is defined by
$$(i,j)(k,l) = \begin{cases}(i,l) & \text{if } j=k \\ 0 &\text{if }j \neq k.\end{cases}$$
where $0$ acting as a zero element in $B_n$. This semigroup is called Brandt Semigroup.
The Endomorphism monoid $End_0(B_n)$ is isomorphic to the monoid $(S_n)^0$ obtained by adjoining the zero map to the group $S_n$, where $S_n$ is the symmetric group of order $n$ and $End_0 (B_n) = \{ \theta \in End(B_n) : 0\theta = 0\}$. A permutation $\sigma \in S_n$ induces the endomorphism of $B_n$ mapping $(i,j) \mapsto (i\sigma , j\sigma)$ and $0 \mapsto 0$
Let $\theta \in End_0(B_n)$. Then $0\theta = 0$. Suppose that for some $(i,j) \in B_n$, we have $(i,j) \theta = 0$. Then for any $p,q \in [n]$,
$$ (p,q)\theta = ((p,i)(i,j)(j,q))\theta = (p,i)\theta (i,j)\theta (j,q)\theta = 0$$
Therefore for any $\theta \neq 0$, we have $(i,j)\theta \neq 0$ for all $i,j \in [n]$
A non zero $\theta \in End_0(B_n)$, then $\exists$ functions $\theta_1, \theta_2 : [n] \times [n] \rightarrow [n]$ such that $(i,j)\theta = ((i,j)\theta_1, (i,j)\theta_2) \ \ \ \ \ \ \ \ \ \ (1)$
Now for any $k \in [n]$, we have $(i,k)\theta (k,j) \theta \neq 0$ iff $(i,k)\theta_2 = (k,j)\theta_1$ and then $(i,k) \theta (k,j)\theta = ((i,k)\theta_1 , (k,j)\theta_2) \ \ \ \ \ \ \ \ (2)$
From eq (1) and (2) ,we get
$$ (i,j)\theta_1 = (i,k)\theta_1$$
$$(i,j)\theta_2 = (k,j)\theta_2$$
It follows that $\theta_1$ depends only on the first cordinates , $\theta_2$ depends only on the second cordinates. Since $\theta$ is non zero, SO $(i,k)\theta (k,j) \theta \neq 0$ , then $(i,k)\theta_2 = (k,j)\theta_1$ for all $k\in [n]$ , then $\theta_1 = \theta_2$
Define $ :\sigma [n] \rightarrow [n]$ such that $i\sigma = (i,j)\theta_1 = (j,i)\theta_2$ , where $j$ is any number in $[n]$ . $\sigma$ is well defined , since $\theta_1$ is a function.
How to show that $\sigma $ is injective.
Any help would be appeciated . Thank you
It follows from (1) that $(i,j)\theta = ((i,j)\theta_1, (i,j)\theta_2) = (i\sigma,j\sigma)$. Suppose that $i\sigma = k\sigma$ for some $k \not= i$. Then \begin{align} (i,i)\theta &= ((i,i)(i,i))\theta = (i,i)\theta (i,i)\theta = (i\sigma,i\sigma)(i\sigma,i\sigma)\\ &= (i\sigma,k\sigma)(i\sigma,i\sigma) = (i,k)\theta (i,i)\theta = ((i,k)(i,i))\theta = 0\theta = 0 \end{align} a contradiction.