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NB: This part of Exercise 1.9.2 of "Fundamentals of Semigroup Theory," by John Howie. There is a similar question here but it doesn't ask for the specific definition below.

The Definition:

Definition: A semigroup $S$ is a group if there exists an $e$ in $S$ such that for all $a$ in $S$, $ea=a$ and for all $x$ in $S$ there exists an $y$ in $S$ such that $yx=e$.

The Problem:

Let $S$ be a semigroup. Show that $S$ is a group (as defined above) if and only if for all $a$ in $S$, $aS=S=Sa$.

Thoughts:

Fix $a\in S$. It is enough to show that both $\lambda_a: S\to S$, given by $s\mapsto as$, and $\rho_a: S\to S$, given by $s\mapsto sa$, are bijections.

Subproblem:

How do I show that $xe=x$ for any $x$ in $S$ given the definition above?

My Attempt:

I'm stumped.

Shaun
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1 Answers1

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Since $aS = S = Sa$ for all $a \in S$, all elements of $S$ are both $\mathcal{R}$- and $\mathcal{L}$-equivalent. It follows that $S$ consists of a single $\mathcal{H}$-class. This $\mathcal{H}$-class contains the product of (any) two of its elements and hence it is a group by Theorem 2.2.5 of Howie's book.

EDIT. This answer to this question answers your question. Indeed, it is shown in this answer that there exists a right identity in $S$. Of course a dual proof would show the existence of a left identity $e$. Now, since for all $x \in S$, $e \in xS$, there exists $y$ such that $yx = e$.

J.-E. Pin
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    Thank you. Whilst this answers the question, I'm looking for something a little more elementary than this proof. (I'm reading Howie's book and I don't want any anachronisms, citing later theorems for earlier exercises.) – Shaun Apr 15 '17 at 15:45
  • I might be able to understand this better once I've read the relevant section. – Shaun Apr 15 '17 at 15:51
  • The proof of Theorem 2.2.5 of Howie's book makes use of (1.1.7) of the book, which states that "a non-empty subset $T$ of $S$ is a subgroup of $S$ if and only if $(\forall a\in T) aT=T$ and $Ta=T$", which uses the result we're trying to prove. – Shaun May 01 '17 at 16:23
  • For completeness, here is Theorem 2.2.5: "If $H$ is an $\mathcal{H}$-class in a semigroup $S$ then either $H^2\cap H=\emptyset$ or $H^2=H$ and $H$ is a subgroup of $S$." – Shaun May 01 '17 at 16:27