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

There's a theorem of Kaplansky that states that if an element $u$ of a ring has more than one right inverse, then it in fact has infinitely many. I could prove this by assuming $v$ is a right inverse, and then showing that the elements $v+(1-vu)u^n$ are right inverses for all $n$ and distinct.

To see they're distinct, I suppose $v+(1-vu)u^n=v+(1-vu)u^m$ for distinct $n$ and $m$. I suppose $n>m$. Since $u$ is cancellable on the right, this implies $(1-vu)u^{n-m}=1-vu$. Then $(1-vu)u^{n-m-1}u+vu=((1-vu)u^{n-m-1}+v)u=1$, so $u$ has a left inverse, but then $u$ would be a unit, and hence have only one right inverse.

Does the same theorem hold in monoids, or is there some counterexample?

• I don't understand your proof. Where do you use the fact that $u$ has more than one right inverse? Commented Jun 9, 2012 at 21:30
• Sure, I'll add it. Maybe it's incorrect. Commented Jun 9, 2012 at 21:31
• Kaplansky theorem is a known fact in semigroup theory. You can find more details about it in the following journals.impan.pl/cgi-bin/doi?sm168-2-7 Commented Apr 6, 2014 at 5:33
• @QiaochuYuan The proof given by Camilla is correct. One element with a right inverse having more than one right inverse is equivalent to that element not being unit, which has been made of use in the last part of proof. Commented Jan 25, 2019 at 2:32

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.

• Nice! I considered this example but somehow convinced myself that $f$ had infinitely many right inverses... Commented Jun 9, 2012 at 21:53
• Thanks! In this semigroup, an element can have either one or infinitely many left inverses. (Or none.)
– user23211
Commented Jun 9, 2012 at 21:55
• Thanks ymar, this is a nice, concrete example. Commented Jun 9, 2012 at 21:58

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 $1 = xy = x^2 y = x$.