The example given in your table is a unital magma. It is not an idempotent magma, as $bb = a ≠ b$ and it is not a semigroup as $(aa)b = b ≠ 1 = aa = a(ab)$. Thus this example solves your first two questions.
Here is an example of an idempotent magma which is not a semigroup:
$\begin{array}{c|rrrr}
& a & b & c \\
\hline
a & a & a & b \\
b & b & b & b \\
c & c & c & c
\end{array}$
If you want an idempotent magma with identity, just add an identity
$\begin{array}{c|rrrrr}
& 1 & a & b & c \\
\hline
1 & 1 & a & b & c \\
a & a & a & a & b \\
b & b & b & b & b \\
c & c & c & c & c
\end{array}$
Briefly, the answer is no: associative quasigroups are always inverse semigroups, but inverse semigroups are not necessarily associative quasigroups.
However, associative quasigroups are equivalently cancellative inverse semigroups!
To be concrete, I'll work with the following definitions:
An associative quasigroup is a set $Q$ with a multiplication $*:Q\times Q\to Q$ such that
- it is associative in that $a*(b*c)=(a*b)*c$
- it has the Latin square property, meaning for $a,b\in Q$ there exists a unique $l$ and a unique $r$ such that
On the other hand:
An inverse semigroup is a set $S$ with a multiplication $\cdot:S\times S\to S$ such that
- it is associative in the same sense that $a\cdot(b\cdot c)=(a\cdot b)\cdot c$
- every element $x\in S$ has a unique inverse $y$ such that
- $x=x\cdot y\cdot x$
- $y=y\cdot x\cdot y$
by the uniqueness, we may denote the inverse $y$ of $x$ by $x^{-1} := y$ in this case
So the question is: are they the same thing?
If $(Q,*)$ is an associative quasigroup and $x\in Q$, then we want to see if it has an inverse $y$ such that $x=x*y*x$ and $y=y*x*y$. What would be a candidate?
By the Latin square property with $a=b=x$, there is a unique $l$ such that $l*x=x$.
Then by the Latin square property again with $a=x$ and $b=l$, there is a unique $r$ such that $x*r=l$.
Therefore, $x=l*x=x*r*x$, so $y=r$ is a candidate for an inverse!
Does it work? First, we need to see if $r*x*r=r$, which might not look apparently true, but it has to do with the fact that quasigroups have the cancellation property. In this particular example, we reason as follows:
First note that $(l*l)*x=l*(l*x)=l*x=x$. Since $l$ is the unique element with the property that $l*x=x$, this means $l*l=l$.
- in other words, given the equation $(l*l)*x=l*x$, we can "cancel out" the $x$ and conclude that $l*l=l$
Now, this means $x*(r*x*r)=(x*r)*(x*r)=l*l=l$. However, $r$ is the unique element such that $x*r=l$, so again we get to conclude from $x*(r*x*r)=x*r$ that $r*x*r=r$, as desired!
Now we need to know that $r$ is the unique inverse of $x$.
Therefore, suppose $y\in Q$ is another inverse of $x$, so that $x=x*y*x$ and $y=y*x*y$.
We want to conclude that $y=r$.
We can once again use the cancellation property:
- $(x*y)*x=x$, but $l$ is the unique element with $l*x=x$, so $x*y=l$
- $x*y=l$, but $r$ is the unique element with $x*r=l$, so $y=r$, completing the proof.
Conclusion: If $Q$ is an associative quasigroup, then it is also an invertible semigroup.
Now for the converse: suppose $S$ is an invertible semigroup, then will it necessarily be an associative quasigroup? Unfortunately, the answer is no, and a counterexample is the following: let $S = \{0,1\}$ with the multiplication operation given by the usual multiplication of numbers.
- $S$ is associative because usual multiplication is associative, so $S$ is a semigroup
- $S$ is actually inverse, surprisingly: the unique inverse of any element is itself
- for $0$, the inverse cannot be $1$ because $1\cdot0\cdot1=0\neq1$
- for $1$, the inverse cannot be $0$ for the same reason
- since $0\cdot0\cdot0=0$ and $1\cdot1\cdot1=1$, it follows that $0$ is its own inverse, and $1$ is its own inverse
However, $S$ is not an associative quasigroup: try the Latin square property with $a=0$ and $b=1$. We would need to find some $r\in S$ such that $a\cdot r=b$. That is, we are trying to solve $0\cdot r=1$.
The left hand side, however, is always zero, so no such $r$ can exist, and therefore $S$ fails to satisfy the Latin square property!
Conclusion: An inverse semigroup is not necessarily an associative quasigroup.
The reason for this failure ultimately boils down to the fact that inverse semigroups don't necessarily have the cancellation property like quasigroups do.
If we further assert that our inverse semigroup is cancellative, then we are in luck: for any $a,b\in S$, we can take $l := b\cdot a^{-1}$ and $r := a^{-1}\cdot b$ so that $a\cdot r=b$ and $l\cdot a=b$.
I'll show why this works for $r$. First, uniqueness is clear: if $a\cdot r=b$ and also $a\cdot r'=b$, then by the cancellation property, $a\cdot r=a\cdot r'$ will force $r=r'$.
Now to check that $a\cdot r=b$. To do so, we will again use the cancellation property
- if we show $a^{-1}\cdot(a\cdot r)=a^{-1}\cdot b$, then the cancellation property will give us that $a\cdot r=b$
- now, $a^{-1}\cdot(a\cdot r)=a^{-1}\cdot a\cdot r=a^{-1}\cdot a\cdot(a^{-1}\cdot b) = (a^{-1}\cdot a\cdot a^{-1})\cdot b=a^{-1}\cdot b$, using the fact that $a^{-1}$ is an inverse of $a$ and thus $a^{-1}\cdot a\cdot a^{-1}=a^{-1}$. This completes the argument.
Conclusion: Associative quasigroups are the same as cancellative inverse semigroups; that is, $(A,\bullet)$ is an associative quasigroup if and only if it is a cancellative inverse semigroup
Best Answer
This paper, for example, uses the terms
for properties defined in two different ways. The second one means a magma where each element has a unique inverse element.
Also, beware because the finite groupoid as in the Cayley table in your question is definitey not cancellative: $$1*0=1*2=2*1=0*1,$$ but $2\ne 0$.