My Math Forum Functoriality of taking the Automorphism group.

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 November 23rd, 2011, 04:29 PM #1 Newbie   Joined: Nov 2011 Posts: 1 Thanks: 0 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 $F:\ \mathbf{Grp}\ \longrightarrow\ \mathbf{Grp}$ such that $F(G)=\operatorname{Aut}(G)$ for all $G\in\mathbf{Grp}$. Let $\mathbf{Inn}$ be the subcategory* of $\mathbf{Grp}$ of groups having no outer automorphisms. More concisely, the groups $G\in\mathbf{Grp}$ satisfying $\operatorname{Aut}(G)\stackrel{\sim}{=}G/Z(G)$, where $Z(G)$ is the center of $G$. Then the restriction of $F$ to $\mathbf{Inn}$ is again a functor*, which is equivalent* to a functor $F'$ satisfying $F'(G)=G/Z(G)$ for all $G\in\mathbf{Inn}$. I am able to prove that no functor satisfying $F(G)=Z(G)$ exists on $\mathbf{Inn}$, by considering the images of the groups $C_3$ and $S_3$ and their natural injection and projection. Now I would think that the fact that taking $Z(G)$ is not functorial implies that taking $G/Z(G)$ 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 $C_3$ and $S_3$ are for taking $Z(G)$. Or of course in any other way you see fit. Thanks in advance, Tristan

 Tags automorphism, functoriality, group, taking

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