# Conditions for Group given Semigroup with Idempotent Element

Another exercise in Seth Warner's "Modern Algebra" (1965). This one is Exercise 7.15.

Let $$(S, \circ)$$ be a semigroup.

Let $$(S, \circ)$$ have an idempotent element $$e$$, that is, such that $$e \circ e = e$$.

Suppose that for all $$a \in S$$:

• there exists at least one $$x \in S$$ such that $$x \circ a = e$$
• there exists at most one $$y \in S$$ such that $$a \circ y = e$$.

Then $$(S, \circ)$$ is a group.

I have started by exploring things like: $$x \circ a = (x \circ a) \circ (x \circ a) = x \circ (a \circ x) \circ a$$

but although I "know" $$e = x \circ a = a \circ x$$ is my identity, I can't actually work out how to prove it.

The question suggests using the alternative axioms for a group: that $$(3)$$ there exists a left identity (rather than "there exists an identity") and $$(4)$$ that every element has a left inverse (rather than "every element has an inverse"), in addition to $$(1)$$ closure and $$(2)$$ associativity.

I have tried exploring $$x \circ e = e$$ to try and prove that $$e$$ is a right identity, but while we know this holds for the specific $$x$$ I can't see how to generalise for all elements of $$S$$.

In problems of this kind, there is usually an ingenious first step that I can't find which agglomerates $$5$$ or so elements, like $$a \circ e \circ x \circ y \circ a$$ or something, from which an obvious conclusion emerges, but I can't find anything that goes anywhere.

Where do I start?

I will use juxtaposition for the operation. Let $$a$$ be any element.
If $$x$$ is such that $$xa=e$$, then $$xae=ee=e=xa$$, so by the second condition applied to $$x$$, we have that $$ae=a$$. This holds for all $$a$$, so $$e$$ is a right identity.
Again, let $$a$$ be arbitrary, and let $$x$$ be any element such that $$xa=e$$. Then $$e=xa=(xe)a=x(ea)$$ since $$e$$ is a right identity. Again applying the second condition to $$x$$ we conclude that $$ea=a$$. Thus, $$e$$ is also a left identity.
• I don't understand where you say: "by the second condition applied to $x$, we have that $axa= a$." All the second condition says is: there is at most one $y \in S$ such that $ay=e$. Why does that follow? Commented Mar 2, 2022 at 6:54
• @PrimeMover The second condition says: if $b$ is any element of $S$, and we have $br=bt=e$, then $r=t$. That is, the equation $by=e$ had at most one solution for any element $b$ of $S$. Now set $b=x$. Since $xa=e$ and $xaxa=e$, then both $a$ and $axa$ are solutions, so $a=axa$. Commented Mar 2, 2022 at 7:00