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November 23rd, 2011, 04:29 PM   #1
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Functoriality of taking the Automorphism group.

I've been stuck on a single problem for quite a while now, it is part of a problem set which serves as an introduction to category theory, for a course on cohomology and the likes. The question is:

"Does forming the automorphism group induce a functor on the category of groups?"

My intuition tells me no, but I am unable to prove this. My knowledge of group theory has proven insufficient to construct a counterexample. I have tried some rather elaborate constructions to prove that no such functor exists, but my knowledge of category theory too is insufficient to apply it in a proof. We have only been given the definitions of category and functor, both covariant and contravariant. A short sketch of my 'most succesful' attempt to a proof so far is the following:

(Overly tedious parts of the proof have been left out, marked by *.)
Assume that there exists a functor such that for all .
Let be the subcategory* of of groups having no outer automorphisms. More concisely, the groups satisfying , where is the center of . Then the restriction of to is again a functor*, which is equivalent* to a functor satisfying for all .

I am able to prove that no functor satisfying exists on , by considering the images of the groups and and their natural injection and projection. Now I would think that the fact that taking is not functorial implies that taking is not functorial, but I am unable to prove this. This would of course conclude the proof almost immediately.

Regardless of the 'effectiveness' of this attempt to a proof, can anyone help me solve this problem? Either by informing me of a lovely property of functors that I am unaware of, or by giving me a (hint to) a counterexample, like the pair and are for taking . Or of course in any other way you see fit.

Thanks in advance,
Tristan
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