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?

  • $\begingroup$ 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 $\endgroup$
    – Victor
    Oct 30, 2013 at 13:06
  • $\begingroup$ 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. $\endgroup$
    – Victor
    Oct 30, 2013 at 13:14
  • $\begingroup$ 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? $\endgroup$
    – Victor
    Oct 30, 2013 at 13:40
  • $\begingroup$ 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) $\endgroup$
    – Victor
    Oct 30, 2013 at 13:48
  • $\begingroup$ 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. $\endgroup$
    – Victor
    Oct 30, 2013 at 14:19

1 Answer 1


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.

In general, the answer is there can be proper quotients. The free group on two generators is generated as a group by the free monoid on two generators. There are lots of groups generated by two elements which generate a free submonoid, like the free metabelian group or the lamplighter group.

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.

  • $\begingroup$ Ben, what is known about "universal inverse semigroups" for semigroups/(nice type of semigroups)? Is there any analogue to Ore's condition for that setting? $\endgroup$
    – Victor
    Oct 30, 2013 at 13:22
  • $\begingroup$ @Victor, that is a good question. I am not sure what has been done. If memory serves David Rees gave a nice proof that Ore monoids embed in their group of units via inverse semigroups (it is in Lawson's book). He starts with an Ore monoid M and shows it is the R-class of 1 in the inverse hull S and proves M^{-1}M =S. He also proves S is E-unitary and that gives the embedding. I don't remember how much still works without Ore and with weaker cancellation assumptions on M. You might look there first. $\endgroup$ Oct 30, 2013 at 13:53
  • $\begingroup$ Thank you, Ben! I didn't know about this proof by David Rees. $\endgroup$
    – Victor
    Oct 30, 2013 at 14:09

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