Let $S$ be a monoid. On p. xvii of P.M. Cohn's *Free Ideal Rings and Localization in General Rings* (CUP, 2006), one reads that

- an element $u \in S$ is
*regular*if (quote) "[...] it can be cancelled, i.e. $ua = ub$ or $au = bu$ implies $a = b$"; - $S$ is a
*conical*monoid if (quote) "$ab = 1$ implies $a = 1$ (and so also $b = 1$)".

On p. 53, one then reads that $S$ is an *invariant* monoid if each of its elements is invariant, an element $c \in S$ being *invariant* if $c$ is regular and $cS = Sc$ (so in particular, an invariant monoid is cancellative).

Now, with these definitions in mind, Problem 0.9.10 in the same book (p. 58) asks:

Is every invariant conical monoid necessarily commutative?

The problem is numbered "10°" by Cohn, and on p. xiv one reads:

[...] open-ended (or open) problems are marked °, though sometimes this may refer only to the last part; the meaning will usually be clear.

At first, I had mistakenly thought (see the first version of this post) that the problem had a trivial answer, due to the following (flawed) argument:

Start with a semigroup $H$, pick an element $e \notin H$, and let $H^{(e)}$ be the (unique) magma obtained by extending the operation of $H$ to a binary operation on $H^{(e)}$ in such a way that $ex = xe := x$ for every $x \in H^{(e)}$. Clearly $H^{(e)}$ is a monoid (some people would call it an

unconditional unitizationof $H$); and it is cancellative, commutative, or invariant if and only if so is $H$, resp. On the other hand, $H^{(e)}$ is obviously conical (regardless of whether this is the case with $H$). Therefore, if $H$ is a non-commutative invariant monoid (such as the monoid of non-zero elements of a non-commutative skew-field), then $H^{(e)}$ provides a negative answer to Cohn's question.

The issue here is that $H^{(e)}$ is a cancellative monoid if and only if $H$ is a cancellative semigroup without unity: If $H$ is a monoid with identity $1_H$, then $xe = x1_H = x = 1_H x = ex$ for every $x \in H$ (and, by construction, $e \ne 1_H$). So, my (new) question is:

Q.Is anyone aware of any progress on Cohn's problem since 2006?

**Update (12/03/2022).** Cohn's problem has been quickly solved by Pace Nielsen below. A second and, in some sense, much easier solution was communicated by George Bergman to Pace Nielsen and is now the bulk of the accepted answer of this thread.