Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\mathrm{Aut}(M^{\rm gp})$?

share|improve this question
2  
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
1  
@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

1 Answer 1

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.