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. $\endgroup$ May 7, 2015 at 18:27

1 Answer 1


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.

  • $\begingroup$ "It is well known... embedded if and only if cancellative ... being cancellative is obviously a necc. condition but it is not sufficient." - These statements are in contradiction with one another. Please advise us. $\endgroup$ Feb 10, 2022 at 6:57
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    $\begingroup$ @PenAndPaperMathematics You are advised that not every semigroup is commutative. For commutative semigroups, cancellative implies embeddable in a group. For non-commutative, the situation is more complicated. $\endgroup$ Feb 12, 2022 at 5:13

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