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Let $X=(X, \cdot)$ be a semigroup, which satisfies $\forall a,b,c\in X$, $(a\cdot b)\cdot c = a\cdot (b\cdot c)$.

In the definition of monoids, the axiom of identity is (a) below. What happens if you replace the axiom with (b)?

  • (a) $\exists e\in X$, $\forall a\in X$, $ae=ea=a$.

  • (b) $\forall a\in X$, $\exists e\in X$, $ae=ea=a$.

I'm considering an example of (b).

Let $G=\{e, a\}$, $G'=\{e',a'\}$ be copies of a group of order 2 ($e\in G$, $e'\in G'$ are identities) and $X = G\sqcup G'$. And define the operation of a element of $G$ and a element of $G'$ to be $e$. Does $X$ hold (b)?

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    It seems that both (a) and (b) hold for your example; but other semigroup axioms fail (such as associativity). – 2017-02-17

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An example of an associative law for which (b) holds but (a) doesn't is where $X = \mathbf{R}$ and $a \cdot b = \max(a,b)$.