When I met with said friend, one natural question came up: what other semigroups can
we find this way?

The first few simple observations we made were

- If either $A$ or $B$ is the identity matrix ${I}_{2}$ or the zero matrix ${0}_{2}$ the resulting semigroup will contain two elements with an identity or a zero element respectively.
- In general, we can always add ${I}_{2}$ or ${0}_{2}$ to the semigroup generated by $A$ and $B$ and obtain a possibly larger one.
- $A,B$ generate a finite semigroup iff $AB$ is of finite order (in the sense that the set of its powers is finite).
- $AB$ has finite order iff its (nonvanishing) eigenvalue is $+-1$.
- For $A$
of rank $1$
we may assume (by base change) that $A$
is one of the two matrices
$$\left(\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right),\left(\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right)\phantom{\rule{0.3em}{0ex}}.$$

So, as a first approach we thought about the following question.

Thinking about the problem a little and experimenting with Macaulay 2 we ended up with the following classification

** Proposition **
For $A=\left(\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right)$ the
solutions for $B$
being of rank one consist of four one-dimensional families, namely (for
$x\in \mathbb{Q}$)

$$\begin{array}{lll}\hfill {F}_{3}\left(x\right)=\left(\begin{array}{cc}\hfill 0\hfill & \hfill x\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right),{F}_{4}\left(x\right)=\left(\begin{array}{cc}\hfill 0\hfill & \hfill 0\hfill \\ \hfill x\hfill & \hfill 1\hfill \end{array}\right).& \phantom{\rule{2em}{0ex}}& \hfill \end{array}$$ Additionally, we have four special solutions $$\begin{array}{lll}\hfill {G}_{1}=\left(\begin{array}{cc}\hfill -1\hfill & \hfill 1\hfill \\ \hfill -2\hfill & \hfill 2\hfill \end{array}\right),{G}_{2}=\left(\begin{array}{cc}\hfill -1\hfill & \hfill -1\hfill \\ \hfill 2\hfill & \hfill 2\hfill \end{array}\right),& \phantom{\rule{2em}{0ex}}& \hfill \end{array}$$

$$\begin{array}{lll}\hfill {G}_{3}=\left(\begin{array}{cc}\hfill -1\hfill & \hfill 2\hfill \\ \hfill -1\hfill & \hfill 2\hfill \end{array}\right),{G}_{4}=\left(\begin{array}{cc}\hfill -1\hfill & \hfill -2\hfill \\ \hfill 1\hfill & \hfill 2\hfill \end{array}\right).& \phantom{\rule{2em}{0ex}}& \hfill \end{array}$$

We can also describe size and the algebraic structure.

- $A$ with ${F}_{1}$ (${F}_{2}$) generates a right (left) zero semigroup (hence of size $2$, except for $x=0$).
- $A$ with ${F}_{3}$ or ${F}_{4}$ generates a semigroup with $AB$ nilpotent (of size $4$, except for $x=0$, where we have the null semigroup of size $3$).
- $A$ with ${G}_{i}$ generate (isomorphic) semigroups of size $8$. These contain two disjoint right ideals, two disjoint left ideals generated by $A$ and $B$ respectively.

Luckily enough, we get something very similar from our alternative for $A$.

**Proposition**
In case $A=\left(\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right)$ the
solutions for $B$
being of rank one consist of five one-dimensional families namely (for
$x\in \mathbb{Q}$)

As before we can describe size and structure.

- $A$ with ${H}_{1}$ (${H}_{2}$) generates a right (left) zero semigroup (as before).
- $A$ with ${H}_{3}$ or ${H}_{4}$ generates a semigroup with $AB$ nilpotent (as before).
- $A$ with ${H}_{5}$ generates the same $8$ element semigroup (as before).

Finally, it might be worthwhile to mention that the seemingly missing copies of the $8$ element semigroup are also dealt with; e.g. $-{G}_{i}$ generates the same semigroup as ${G}_{i}$ etc.

At first sight it seems strange that we cannot find other semigroups with two generators like this. As another friend commented, there’s just not enough space in the plane. I would love to get some geometric idea of what’s happening since my intuition is very poor. But that's all for today. -->
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