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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?

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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.

Now we have a left identity, and every element has at least one left inverse, so we have a group.

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  • $\begingroup$ 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? $\endgroup$ Commented Mar 2, 2022 at 6:54
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    $\begingroup$ @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 $a=axa$. $\endgroup$ Commented Mar 2, 2022 at 7:00
  • $\begingroup$ Aha, yes thanks, got it, sorry, caffeine needed. Brilliant. $\endgroup$ Commented Mar 2, 2022 at 7:01

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