Under what assumptions can a semigroup $(S,*)$ be embedded into a group?
When a semigroup can be embedded into a group
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0A related (stronger!) question is the following: Let $\langle X; \mathbf{r}\rangle$ be a presentation such that every relator is a positive word, or is (up to cyclic shift) of the form $uv^{-1}$ where $u$ and $v$ are positive words (by "positive word" I mean the word). Such a presentation clearly defines a monoid. Then does the monoid given by the presentation embed in the group given by the presentation? Coxeter groups, for example, have a positive solution to this question. – 2011-11-07
1 Answers
It is clear that if a semigroup is embedded into a group, it must be cancellative.
Clifford, Preston: The Algebraic Theory of Semigroups - Page 34. See also A. Nagy: Special classes of semigroups, Theorem 3.10, p.46
A commutative semigroup can be embedded in a group if and only if it is cancellative. The usual procedure for doing this by means of ordered pairs, is just like that of embedding an integral domain in a field.
This construction is sometimes called group of quotients. This article at proofwiki is somewhat related to this construction.
For non-commutative semigroups, the situation is more complicated.
Clifford, Preston: The Algebraic Theory of Semigroups - Page 36
A cancellative semigroup $S$ can be embedded in a group of left quotients of $S$ if and only if it is right reversible.
This paper might also be of interest: George C. Bush: The embeddability of a semigroup--Conditions common to Mal'cev and Lambek, Trans. Amer. Math. Soc. 157 (1971), 437-448.
EDIT: See also this wikipedia article which contains all the information I've given above and also a few further references. (Perhaps I should have searched wikipedia first, before posting this...)
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0It is also called the group of *differences* when using additive notation. – 2016-05-16