Question about kinda-sorta identities in algebra A friend showed me some semi-groups which were constructed as follows. Let $Q = \{ 1, \ldots, n \}$ be a finite set, and let the operation $\cdot$ be defined on $(Q^{2})^{2}$ by
\begin{align*}
(t(\beta), s(\beta)) \cdot (t(\alpha), s(\alpha)) & = \begin{cases}
(t(\beta), s(\alpha)) & s(\beta) = t(\alpha), \\
\natural & s(\beta) \neq t(\alpha),
\end{cases}
\end{align*}
where $\natural \alpha = \alpha \natural = \natural$ for all $\alpha \in Q^{2}$. Now, there is obviously no "uniform" identity, i.e. no element $\epsilon \in Q^{2}$ for which $\epsilon \alpha = \alpha \epsilon = \alpha$ for every $\alpha$. However, what we can do is make $\epsilon_{j}$, where $\epsilon_{j} = (j, j)$, so $\epsilon_{j} (j, i) = (j, i)$, i.e. a left identity for a particular subset of $Q^{2}$.
My question: Is there a name for these? Moreover, I'm not a very learned algebraist, but I am adept enough with linear algebra. Does this phenomenon occur in matrix multiplication as well, i.e. matrices that will act as left identities for particular families of matrices? If so, what would they look like?
 A: These semigroups are known as aperiodic Brandt semigroups. Your definition is slightly incorrect (since $\sharp$ should be an element of the semigroup). Here is the same definition, with improved notation:
Let $B_n$ be the semigroup with $0$ defined on $\{1, ..., n\} \times \{1, ..., n\} \cup \{0\}$ by setting $0x = x0 = 0$ for all $x \in B_n$ and
$$
(i,j)(k,l) = \begin{cases}
 (i,l) & \text{if $j=k$}\\
 0 & \text{otherwise}
\end{cases}
$$
The semigroup $B_2$ and its monoid version $B^1_2$ (with an identity added) are very famous in semigroup theory. The monoid $B^1_2$ has a nickname: the universal counterexample.
Edit. The semigroup $B_2$ can be described as the set of matrices
$$\left\{  
 \begin{pmatrix} 1&0 \\ 0&0 \end{pmatrix}, 
 \begin{pmatrix} 0&1 \\ 0&0 \end{pmatrix}, 
 \begin{pmatrix} 0&0 \\ 1&0 \end{pmatrix}, 
 \begin{pmatrix} 0&0 \\ 0&1 \end{pmatrix}, 
 \begin{pmatrix} 0&0 \\ 0&0 \end{pmatrix}
 \right\}$$
under usual multiplication of matrices. This can be easily extended to $B_n$.
