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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 $M^{gp}$?2012-06-27
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    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: $M\to M^{gp}$?2012-06-27
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    @ThomasAndrews yes it is.2012-06-27
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    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?2012-06-27
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    @ThomasAndrews every element of $M^{gp}$ induces an element of $Aut(M^{gp})$ by translation. This is a subgroup. Any $a\in Aut(M)$ induces an element of $Aut(M^{gp})$ simply by $a(-x)=-a(x)$. Is it true that $Aut(M^{gp})/M^{gp}$ is in bijective correspondence with $Aut(M)$?2012-06-27
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    Every automorphism of $M$ induces an automorphism of $M^\textrm{gp}$, because $(-)^\textrm{gp}$ is a functor. But I see no reason for the converse to be true.2012-06-27
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    @ullo You need to be precise before people can answer your question.2012-06-27
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    @ullo: Your comment makes no sense to me: the only translation which is a monoid or a group homomorphism is the identity: every automorphism of a monoid/group must map the identity to the identity, so I don't see how you are claiming that $M^{\rm gp}$ maps to $\mathrm{Aut}(M^{\rm gp})$.2012-06-27
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    Is at least the functor $(-)^{gp}$ injective? meaning that $N-> M$ injective implies $N^{gp} -> M^{gp}$ injective2012-06-27
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    @ullo: use `\to` for right arrows, by the way.2012-06-27
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    @ullo That last question isn't even about the automorphism groups. As I've said, you really need to clarify your question.2012-06-27

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