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?



Monoids and groups of fractions
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. 


