When semigroups are groups

A semigroup is a nonempty set $G$ with a binary operation on $G$ that is associative. To be a group, $G$ also needs an identity element and the existence of a two-sided inverse for each element.

Hungerford lists the following propositions that give conditions on when a semigroup is a group:

\begin{prop} Let $G$ be a semigroup. Then $G$ is a group iff the following conditions hold: \begin{enumerate} \item $\exists e \in G, \forall a \in G, ea=a$ \item $\forall a \in G, \exists a^{-1} \in G, a^{-1} a = e$ \end{enumerate} \end{prop} \begin{prop} Let $G$ be a semigroup. Then $G$ is a group iff $\forall a, b \in G$ the equations $ax=b$ and $ya=b$ have solutions in $G$. \end{prop}

Hungerford proves Proposition 1 and leaves as an exercise to show Proposition 2 holds from Proposition 1. This is not as simple as it may first appear. Having solutions of $ax=b, ya=b$ immediately gives elements that are left and right identities for particular elements of $G$, but we need all these identities to be equal to each other to satisfy the conditions of proposition 1. The second condition is achieved trivially.

Proof. Let $a \in G$. Find solutions $x, y \in G$ such that $ax = a, ya=a$. Then find solutions $z, w \in G$ such that $za = y$ and $aw = x$. $$y = za = zax = yx = yaw = aw = x$$ Call this two-sided identity element of $a$, $e_a$. Note that $$a e_a^2 = (a e_a) e_a = a e_a = a$$ So $e_a$ is idempotent. Let $b \in G$ and find solutions $x, y \in G$ to $x e_b = e_a, e_a y = e_b$. $$e_a = x e_b = x e_b^2 = e_a e_b = e_a^2 y = e_a y = e_b$$ ◻

Here is a different characterization of a group:

\begin{prop} A semigroup $G$ is a group iff $\forall x \in G, \exists ! y \in G, xyx = x$. \end{prop}

Proof. For $x \in G$, denote by $x'$ the unique element in $G$ such that $x x' x = x$. Then, $$x (x' x x') x = (x x' x) x' x = x x' x = x$$ Therefore, $x' x x' = x'$, so that $x'' = x$. Also note that $x x'$ must be idempotent because $$(x x')^2 = x x' x x' = x x'$$ For $i \in G$ an idempotent element, $ix = ix (ix)' ix = ix (ix)' i^2 x$ so that $(ix)' = (ix)' i$. Therefore, $$(ix)' = (ix)' (ix)'' (ix)' = (ix)' ix (ix)' = (ix)' x (ix)'$$ Implying $ix = x$. Since $x$ was arbitrary, this means that $i$ is a left identity in $G$. With a symmetric argument, we can prove that $i$ is also a right identity. Since $i$ was arbitrary, all idempotents are equal to each other. In particular, $x x'$ is an idempotent, hence $x'$ is an inverse of $x$. ◻

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