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Does there exist a commutative semigroup $S$ with the following (additively denoted) properties?

  1. For all $x \in S$, the function $S \rightarrow S$ given by $y \mapsto x+y$ is a bijection.
  2. $S$ has no identity element; that is, there is no $0 \in S$ such that for all $x \in S$, $x+0=x.$
  3. $S$ is non-empty. (This is to rule out the trivial example of $S=\emptyset$, thanks to Najib Idrissi for pointing this out.)

Of course, if 2 is replaced by its negation (namely, that there is an identity element), then examples are easy to find; in fact, these are precisely the Abelian groups.

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  • $\begingroup$ I'm pretty sure you meant $x+0=x$ in condition 2, too. $\endgroup$ – Najib Idrissi Dec 7 '14 at 20:13
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Assume $S$ is nonempty, containing the element $x$, and is such that addition of any element gives a bijection. Then since addition of $x$ is a bijection, there exists a $y$ such that $x+y=x$. We claim that $y$ is an identity element. Let $z\in S$. Then, since addition of $x$ is a bijection, there exists a $w$ such that $z=w+x$. But then $z+y=(w+x)+y=w+(x+y)=w+x=z$. Thus, $S$ does indeed have a $0$, and is therefore an abelian group (inverses exist by the bijectivity of addition).

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  • $\begingroup$ Right. So to show that there exists a $0$ element, all we need is a commutative semigroup $S$ together with an $x \in S$ such that $x \mapsto x+y$ is a surjection. Now that's cool. $\endgroup$ – goblin Dec 7 '14 at 23:15
  • $\begingroup$ Yes, addition does not even have to be injective, though if it is surjective, my proof shows that it makes $S$ into an abelian group and is thus injective. $\endgroup$ – Nishant Dec 8 '14 at 0:07
  • $\begingroup$ I like it. $\;\!$ $\endgroup$ – goblin Dec 8 '14 at 1:20
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No. Let $S \neq \emptyset$ be as in your question. Let $a \in S$ be arbitrary, $T_a : S \to S$ be defined by $T_a(x) = a + x$ and let $0 = T_a^{-1}(a)$. Then for all $x \in S$, $$T_a(x+0) = a + x + 0 = x + (a+0) = x + T_a(T_a^{-1}(a)) = x + a = T_a(x)$$ Since $T_a$ is a bijection, $x+0=x$, and this is for any $x \in S$.

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