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A semigroup is a set $S$ together with an associative binary operation $m:S\times S\rightarrow S$.

In any kind of semigroup I can think of (group, ring, field, etc), this binary operation $m$ is not invertible.

If $|S|<\infty$, then $m$ cannot be invertible. Edit: Wrong! See accepted answer.

I would be interested in an example of semigroup such that $m$ is invertible... or a proof that no such object exists.

In such a semigroup, unless I am being an idiot, associativity is equivalent to

$$m\times \operatorname{id}_S=\operatorname{id}_S\times m.$$

I was going to say this is a kind of commutativity. Now I realise this is nonsense... but I still think it might be interesting to come up with a semigroup with invertible $m$.

You may assume the axiom of choice.

JP McCarthy
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  • @bof I had originally written this with magmas in mind: than I deleted it and rewrote it for semigroups but forgot to mention associativity... fixed now thank you. – JP McCarthy Sep 28 '19 at 18:23
  • @bof there are loads, are there not? For example $g^{-1}$ here (https://math.stackexchange.com/a/325241/19352) as the binary operation and $\mathbb{N}$ the magma. – JP McCarthy Sep 28 '19 at 18:29

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It is not true that if $|S|<\infty$ then $m$ cannot be invertible: in fact, for the cardinalities to match we have $|S|\times |S|=|S|$, meaning that $|S|$ is a root of $x^2-x=x(x-1)$ so $|S|=1$ or $|S|=0$ and both of these possibilities result in (trivial) semigroups.

Now I will show that these are the only two possibilities even without the assumption $|S|<\infty$. Since $m$ is injective, for any $a,b,c,d\in S$ with $m(a,b)=m(c,d)$ we can conclude that $a=c$ and $b=d$. Recall the semigroup property $$ m(m(a,b),c)=m(a,m(b,c)). $$ Thus, $m(a,b)=a$ and $m(b,c)=c$ by injectivity. But $a,b,c$ are arbitrary, so if we choose $b=c$ in the first equation we get $m(a,c)=a$ and if we choose $b=a$ in the second equation we get $m(a,c)=c$, which means that $a=c$ must hold for arbitrary $a,c\in S$, which means that $|S|\leq 1$ and we get the same two possibilities as in the finite case above.

pre-kidney
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