# Does this particular axiom on a semigroup guarantee that it is a group?

Update: Eric Wofsey has demonstrated the conjecture in the commutative case below, and Tobias Kildetoft has provided a simple counterexample to the non-commutative claim.

This would-be replacement for the usual group axioms was suggested by the ##math IRC channel user Aleric, and no solution has been found so far. In its original form, the conjecture reads:

Let $(S,+)$ be a non-empty commutative semigroup satisfying the following "reversibility" axiom: for all $x,y \in S$, there exists a $z \in S$ such that $$x + y + z = x.$$ Then there exists an identity, i.e., an element $0 \in S$ such that $x + 0 = x$ for all $x \in S$.

Of course, if this is true then the identity must be unique, and $S$ becomes an Abelian group by applying the reversibility axiom to $0,x$.

I suppose one could just as easily drop the commutativity condition and formulate a stronger conjecture as follows:

Let $(S,\cdot)$ be a non-empty semigroup satisfying the following "reversibility" axiom: for all $x,y \in S$, there exist $z,w \in S$ such that $$xyz = wyx = x.$$ (Actually, it's not clear to me if we shouldn't ask for a sole element $z = w$ instead.)

Then there exists an identity, i.e., an element $1 \in S$ such that $x1 = 1x = x$ for all $x \in S$.

Again $S$ becomes a group by applying the reversibility axiom to $1,x$.

I have checked all the commutative semigroups of order $3$ and found none which satisfy this axiom and aren't groups, but this isn't very satisfactory. I am looking for a proof of either conjecture or a counterexample.

• For the noncommutative case you can get a counter example by letting all products equal the left-most element. – Tobias Kildetoft Jul 23 '16 at 21:33
• @TobiasKildetoft Ah, yes, of course! Thank you! – Alex Provost Jul 23 '16 at 21:45

In the commutative case, as long as $S$ is nonempty it has an identity (of course, $S=\emptyset$ is a counterexample to your conjecture as originally stated). Indeed, let $x\in S$; then there exists $y$ such that $x+x+y=x$. Let $0=x+y$, so $x+0=x$. Let $z\in S$ be arbitrary; we wish to show $z+0=z$. Note that there exists $w$ such that $z+x+w=z$. Thus $$z+0=z+x+w+0=z+w+(x+0)=z+w+x=z.$$ For the sake of completeness, let me include the counterexample in the noncommutative case that Tobias Kildetoft gave in the comments. Given any set $S$, you can make $S$ a semigroup by defining $xy=x$ for all $x,y\in S$. This will not have a unit unless $S$ has one element, but will satisfy your condition since you can take any $z$ and $w=x$.