A commutative monoid in which cancellation holds need not be a group: $(\mathbb{N}, +)$

False claim: every commutative monoid (Semigroup and monoid) in which the cancellation law holds, that is in which $x + z = y + z$ implies $x = y$, is a group (Group and abelian group).

The natural numbers under addition refute it: $(\mathbb{N}, +, 0)$ is a commutative monoid, cancellation holds in it, and it is not a group, because $1$ has no additive inverse.

• This is the sharpest available refutation, not merely a refutation. The weaker observation that $(\mathbb{N},+)$ is not a group leaves open the possibility that cancellation is what is missing; the point here is that cancellation is present and still does not suffice. In a group cancellation is a theorem (Cancellation in a group: $gx = gy$ or $xg = yg$ forces $x = y$; equivalently left and right translation by $g$ are bijections of $G$, so $gx = h$ and $xg = h$ each have exactly one solution), so the implication runs one way only.

• What is missing is exactly invertibility, and the standard remedy is to adjoin it: the construction of $\mathbb{Z}$ from $\mathbb{N}$ (The integers as equivalence classes of pairs of naturals) is precisely the passage from this cancellative monoid to a group containing it, which is why the pairs $(a,b)$ there stand for formal differences.

• The element $1$ is not special: no $n \ne 0$ has an additive inverse in $\mathbb{N}$, by the same computation with $n$ in place of $1$ once $n$ is written as a successor. One witness suffices to refute the claim.