Kaplansky's theorem of infinitely many right inverses in monoids?

Solution 1:

Let $S$ be the semigroup of functions from $\mathbb N=\{z\in \mathbb Z|z\geq 0\}$ to itself, with the composition written traditionally: $(f\circ g)(x)=f(g(x)).$

Let $f\in S$, $f(0)=f(1)=0$ and for $n\geq 2,\,f(n)=n-1.$ Suppose $f\circ g=\operatorname{id}$. Then for $n\geq 1$, we must have $g(n)=n+1$. However, $g(0)$ can be chosen to be either $0$ or $1$ and the equality holds.

Solution 2:

No. The free counterexample is the monoid $M$ generated by three elements $x, a, b$ where $xa = xb = 1$ and we impose no other relations. Its elements can concretely be described as words on the alphabet $\{ x, a, b \}$ in which $x$ never appears to the left of $a$ or $b$ with the obvious composition and in this monoid $a$ and $b$ are the only right inverses of $x$. (The point is that each element of the monoid has a unique normal form of minimal length and it is straightforward to find a normal form of the product of $x$ and another element in normal form; if the other element starts with $a$ or $b$ then we cancel it and otherwise we can do no further canceling.)

My first attempt to write down a counterexample failed very badly:

• It was commutative. This can't work because right inverses are left inverses in a commutative monoid.
• It was finite. This can't work because finite monoids acts faithfully on finite sets (namely themselves), so right inverses are also left inverses in this case.
• Every element in it was idempotent. This can't work because non-identity idempotents are never right-invertible: if $x^2 = x$ and $xy = 1$, then $1 = xy = x^2 y = x$.