Surjectivity of right and left translations implies semigroup is in fact group

Question

I want to show a set of statements on a semigroup $G$ are equivalent. The left and right translations are given by $l_g(h)=gh$ and $r_g(h)=hg$ respectively.

$G$ is a group
For all $g \in G$ both $l_g$ and $r_g$ are bijective
For all $g \in G$ $r_g$ is surjective and there exists an $f\in G$ s.t. $l_f$ is surjective

The implications from 1. to 2. and 2. to 3. I managed. But I'm struggling with proving 3. implies 1. I think I only need to show the existence of the neutral and inverse elements, as $G$ is by assumption a semigroup and thereby closed.

A neutral element for each element must be in $G$ as $r_g$ is surjective so in particular $\forall g\in G: \exists e\in G$ s.t. $g=r_g(e)$. How can I show that this e is the same for all $g$?

I still don't see how I can show the inverse elements lie in $G$

You’ve shown that for each $g\in G$ there exists $e\in G$, which may depend on $g$, with $eg=g$. You have not shown that the same $e$ works for every $G$, so you have not shown that $G$ has a neutral element. — Arturo Magidin, Apr 20 '20 at 21:46
@ArturoMagidin even though it's not explicitly stated, I think they implicitly don't allow the empty semigroup. Good point though, I didn't even think of that counterexample. — Ruben Kruepper, Apr 20 '20 at 22:03
A lot of authors disallow the empty semigroup; that would satisfy 2, but neither 1 nor 3. — Arturo Magidin, Apr 20 '20 at 22:04
@ArturoMagidin the exercise asks us to prove the statements, so I believe it was an oversight on part of the author. — Ruben Kruepper, Apr 20 '20 at 22:16
Or the author may simply have defined “semigroup” so as to exclude the empty set. That’s fairly common. — Arturo Magidin, Apr 20 '20 at 22:18

score 1 · Accepted Answer · answered Apr 20 '20 at 22:47

1

(From the statement, it seems clear that the particular author does not consider the empty set to be a semigroup; this does not matter for the implication $3\Rightarrow1$, as the statement of 3 implies $G$ is nonempty, but it affects the proof of $2\Rightarrow 3$.)

Let $e_f$ be such that $e_ff=f$. For every $g\in G$, there exists $x$ such that $fx=g$; therefore, for every $g\in G$ we have $$e_fg = e_f(fx) = (e_ff)x = fx = g.$$ Thus, $e_f$ is a left identity for $G$.

And for every $g\in G$ there exists $x$ such that $xg=e_f$. Thus, $G$ has a left identity and every element has a left inverse. These two conditions are known to guarantee that a semigroup is in fact a group.

answered Apr 20 '20 at 22:47

Arturo Magidin

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For future readers: Here is a proof of the fact that these two conditions guarantee a semigroup is a group. https://www.youtube.com/watch?v=E1WNqKSDGio – Ruben Kruepper Apr 21 '20 at 08:12
@RubenKruepper: The problem has been solved in this site; that one is for right identity and right inverse, but the solution is the same mutatis mutandis. – Arturo Magidin Apr 21 '20 at 15:25

Surjectivity of right and left translations implies semigroup is in fact group

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