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What conclusions can be drawn about the relations between two objects with the same group of endomorphism?

Can we tell from End(A) if A is Abelian or not?

Does End(A) contain information about the sub-objects of A?

Any information or references to information about this is highly appreciated.

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So $A$ is a group? –  Rasmus Dec 11 '10 at 18:49
    
Groups, Rings, Modules.. Any algebraic structure over which a group or ring of endomorphisms make sense. I saw from the book Wagner linked to that these include categories, but alas I am not that advanced yet. –  Feydarkin Dec 11 '10 at 19:18
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End(A) can also tell you about whether or not A splits as a direct sum, which is fun to know. –  Sean Tilson Dec 11 '10 at 21:56
    
If "groups" are the things where you can multiply and divide, then "additive categories" are just the things where End(A) is a ring. Nice additive categories have the property that their objects are modules, and End(A) is exactly the ring of module endomorphisms. Probably for you then, you can just stick to modules. –  Jack Schmidt Dec 11 '10 at 22:20
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2 Answers

up vote 5 down vote accepted

For abelian groups, the ring End(A) is very important. As far as non-abelian groups A go, End(A) is not even (usually considered) a group.

"Adding" homomorphisms doesn't work in the non-abelian case.

If you define (f+g)(x) = f(x) + g(x), then (f+g)(x+y) = f(x+y) + g(x+y) = f(x) + f(y) + g(x) + g(y), but (f+g)(x) + (f+g)(y) = f(x) + g(x) + f(y) + g(y). To conclude that:

    f(y) + g(x) = g(x) + f(y)

are equal, you use that + is commutative, that A is abelian. More precisely, if you take f=g to be the identity endomorphism, then f+g is an endomorphism iff A is abelian.

"Composing" homomorphisms doesn't work to form a group, since they are not invertible.

Aut(A), the group of invertible endomorphisms, does form a group. Aut(A) does not determine if a group is abelian or not: 4×2 and the dihedral group of order 8 have isomorphic automorphism groups.

Instead of a ring, End(A) sits inside the "near-ring" of self-maps. See the wikipedia article on nearring for an explanation.

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Thank you, that was a really nice answer. –  Feydarkin Dec 11 '10 at 23:05
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Here is a partial answer (although it uses the ring structure on the set of endomorphisms). A torsion abelian group is cyclic if and only if any two elements in $End(A)$ commute with respect to composition (in other words, $End(A)$ is a commutative ring).

http://www.springerlink.com/content/m72222448q6j7327/

There is a whole book on this subject: Endomorphism rings of abelian groups.

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Thank you for the answer. Do you know if anything useful can be said about non-abelian structures as well. It seemed to me like the book you referenced focuses on commutative structures alone. –  Feydarkin Dec 11 '10 at 19:20
    
@Feydarkin: well, the title tells you so! –  Mariano Suárez-Alvarez Dec 11 '10 at 20:07
    
@Feydarkin: Sorry, I did not have anything useful to add, but I think Jack Schmidt answers your question very well. –  Timothy Wagner Dec 11 '10 at 23:07
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