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If a semigroup $S$ has a left identity-element, then for any $y \in S$ we can write $y = xy$ for some $x \in S$. Just take $x$ to be any of the left identities, of which there is at least one, by hypothesis.

Does there exist a semigroup $S$ such that every $y \in S$ factorizes in this way, which nonetheless lacks a left identity-element?

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Every idempotent semigroup has this property, because $y=yy$, but not all have left identities.

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  • $\begingroup$ Ah but of course. $\endgroup$ – goblin Nov 19 '13 at 17:41
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To give a concrete example based on vadim123's answer:

Let $X$ denote an arbitrary set (for ease of imagining, assume non-empty). Then $2^X$ can be made into an idempotent monoid by defining composition as binary union. Now delete the empty set from $2^X$, obtaining a semigroup $S$. Since $S$ is idempotent, thus every $A \in S$ factorizes as $A \cup A$. Nonetheless, $S$ has no identity element.

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    $\begingroup$ Another example would be $\mathbb{Z}$ with the binary operation $\max$. $\endgroup$ – Najib Idrissi Nov 19 '13 at 18:05

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