# automorphism of a commutative monoid

let $M$ be a finitely generated, commutative monoid.

What are in general the relations between $\mathrm{Aut}(M)$ and $\mathrm{Aut}(M^{\rm gp})$?

When is it true that $\mathrm{Aut}(M)$ determines $\mathrm{Aut}(M^{\rm gp})$?

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What is $M^{gp}$? – Marc van Leeuwen Jun 27 '12 at 15:56
Is $M^{gp}$ the adjoint of the forgetful function from the category of groups to the category of monoids? That is, is $M^{gp}$ the universal group with a corresponding (monoid) homomorphism: $M\to M^{gp}$? – Thomas Andrews Jun 27 '12 at 16:09
@ThomasAndrews yes it is. – ullo Jun 27 '12 at 16:16
Also, what do you mean by "determines?" Do you mean if $M$ and $N$ are monoids, and you know that $Aut(M)\cong Aut(N)\cong G$, under what conditions on $G$ can we prove that $Aut(M^{gp})\cong Aut(N^{gp})$? Or are you asking something else? – Thomas Andrews Jun 27 '12 at 16:23
@ullo That last question isn't even about the automorphism groups. As I've said, you really need to clarify your question. – Thomas Andrews Jun 27 '12 at 17:36

I don't think there's much to say in general. For example, if every element of $M$ is idempotent ($m^2 = m$) then $M^{gp}$ is trivial. To generate examples of this you can take any finite lattice with join as the monoid operation.
Of course since $M^{gp}$ is a functor every automorphism $M \to M$ induces an automorphism $M^{gp} \to M^{gp}$. But my point is this need not be injective.