# Monoids and groups of fractions

Let $G$ be a group containing a monoid $M$ that spans $G$ as a group. Is it possible to have a proper quotient $\varphi \colon G \to Q$ of $G$ such that the restriction of $\varphi$ to $M$ is injective?

More specifically I'm interested in the following: if $M$ is an Ore monoid (cancellative and admitting least common right multiples) then it embeds into its group of right fractions $Q$. There is also a universal group $G$ through which any map from $M$ to a group factors (it has presentation $\langle M \mid m \cdot n = (mn) \text{ for }m,n \in M\rangle$). So $Q$ is a quotient of $G$. Can it be a proper quotient?

In other words: $Q$ is by definition universal among the groups $\iota \colon M \to H$ subject to the condition that $H = \iota(M) \cdot \iota(M)^{-1}$. Is it nontheless universal among all groups?

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From the work of Patrick Dehornoy "Complete positive grop presentations" it follows nicely, using rewriting systems (brilliant tool indeed!), that the group of quotients for Ore's monoids is universal. You may also enjoy Example 23 from a recent preprint by George Bergman arxiv.org/abs/1309.0564, as well as the whole paper – Victor Oct 30 '13 at 13:06
Excuse me to mention also this: Baumslag-Solitar monoids $\langle a,b:ab=ba^k\rangle$ have Ore's condition, but how much different is the behaviour of the BS-monoid an dthe corresponding BS-group! In general it seems very little in common between a group-embeddable semigroup and its universal group. – Victor Oct 30 '13 at 13:14
Does anybody know any instances in the past when Ore's monoids would appear naturally in order to do something good. I know only of Grigorchuk's proof that f.g. cancellative semigroups with polynomial growth are exactly virtually nilpotent f.g. cancellative semigroups? – Victor Oct 30 '13 at 13:40
Sorry, i recalled now: you may check with Simon Craik that (f.g.) Ore's monoids sometimes may be nice -- say they admit only 1,2 or continuum many ends, but i think it is still unknown whether Stallings theorem for them could be worked out (by ends i mean the more natural undirected ends rather than care about artficial in semigroups setting directions in Cayley graphs) – Victor Oct 30 '13 at 13:48
Sorry again, i ate biscuits so cannot stop: it seems would be really nice to understand what are the group-embeddable semigroups whose universal groups are one-relator. At least what are those with Ore's condition. You see, i'm trying to crystalise "importance" of group-embeddable semigroups, which so far only show that things are wild. – Victor Oct 30 '13 at 14:19

For an Ore monoid the universal group is the group of right fractions. This is proved just as the universal property for localization of commutative rings is proved. It is irrelevant whether $H=\iota(M)\iota(M)^{-1}$, you can simply send a fraction $(m,n)$ to $\iota(m)\iota(n)^{-1}$ and check that this gives a well defined homomorphism. Or you can look in a category theory book for the words "calculus of right fractions" where they will prove a more general universal property for localizing categories. Think of a monoid as a one object category and their condition that the monoid admit a calculus of right fractions is the Ore condition.
Edit: Alternatively you can show that if $M$ satisfies an Ore condition and $\iota\colon M\to H$ is a homomorphism to a group $H$ then $\iota(M)\iota(M)^{-1}$ is a subgroup of $H$ and so the two universal properties are the same. It is basically the same argument as to why you can form the product of 2 fractions.