As stated in the answer of Is the forgetful functor from groups to monoids right adjoint? , the forgetful functor $U:\mathbf{Grp}\rightarrow\mathbf{Mon}$ has a left adjoint $G$, and Grothendieck's construction is involved.

Now, my intuition tells me that for a monoid $M$ the following holds:

$M$ can be embedded in some group if and only if $M \longrightarrow G(M)$ is injective.

Is this true? Can we find a characterization of such monoids?

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    $\begingroup$ For commutative or finite monoids, cancellation is a sufficient condition. $\endgroup$ – Slade May 7 '15 at 18:27

Your statement is correct. The universal property of $G(M)$ is that a homomorphism of monoids $M \to G$, where $G$ is a group, extends uniquely to a group homomorphism $G(M)\to G$ via the map $M\to G(M)$.

It follows that if there exists an injective homomorphism $M\to G$, then the map $M\to G(M)$ is injective.

It is well known that a commutative semigroup can be embedded in a group if and only if it is cancellative, and furthermore that every finite cancellative semigroup is a group.

In general, being cancellative is obviously a necessary condition, but it is not sufficient (though I don't have a counterexample on hand). It seems that two different equivalent sets of conditions were worked out by Malcev and Lambek.

Both are discussed in this 1971 paper of Bush, but this seems quite involved, and a more detailed discussion probably belongs on mathoverflow.


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