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Let's call $\frak{G}$ the class of all groups. Let's consider on $\frak{G}$ the equivalence relation $\sim$ such that $G \sim G' \Leftrightarrow \exists \varphi $ isomorphism of groups such that $G=\varphi(G')$

Consider $\chi: (\frak{G}/\sim) \rightarrow (\frak{G}/\sim)$ such that $\chi([G]_{\sim})=[Aut(G)]_{\sim}$.

What can I say about $\chi$ (one-to-one, onto)? What if I just consider finite groups?

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The set of all groups does not exist. You need to say “the class of all groups.” –  Haskell Curry Jan 4 '13 at 16:45
A simpler way of asking this question is "can non-isomorphic groups have isomorphic automorphism groups?" The answer is "yes they can", as illustrated by Andrea Mori below. –  Clive Newstead Jan 4 '13 at 16:47
related: mathoverflow.net/questions/5635/does-autaut-autg-stabilize –  user29743 Jan 4 '13 at 16:50
I think this question is too vague, and you need to ask something more specific. –  Derek Holt Jan 4 '13 at 16:52
You want to know as much as you can about the operation of taking the automorphism group of a group? This is far far far too broad for this site. –  Chris Eagle Jan 4 '13 at 17:32
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3 Answers

up vote 7 down vote accepted

To begin with, there is no such thing as the 'set of all groups.' What you are looking for is a proper class, for which this question should be helpful.

In response to the part about $\chi$ being injective, this is surely not true, as many groups have the same automorphism group - the trivial group and $\mathbb{Z}_2$, for a simple example.

An interesting paper by Iyer contains numerous other examples of groups with the same automorphism groups, in particular Prop. 6.1 - 6.6. He also shows that a nonabelian simple group has the same automorphism group as its covering group.

Furthermore $\chi$ is not surjective. For example, $\mathbb{Z}_p$ is not an automorphism group of any group for $p>2$. (Jack Schmidt says it best here.) Additionally, there is no group $G$ such that $\text{Aut}(G)$ is an Abelian $p$-group of order $\leq p^{11}$ for $p>2$. (reference.) It is also worth noting that Iyer proves that $S_6$ is not the automorphism group of any finite group, and that every finite group occurs as the automorphism group of at most finitely many finite groups.

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$\operatorname{Aut}(\Bbb Z/4\Bbb Z)\simeq\operatorname{Aut}(\Bbb Z/3\Bbb Z)$.

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+1: A simple counterexample says it all. –  Haskell Curry Jan 4 '13 at 16:47
What about onto?! –  Ivan Jan 4 '13 at 16:51
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There's also another thing which can be said about the map above: it cannot be extended to a functor: this is an exercise from Barr and Wells' Topos Triples and Theories. A prove of this fact is given by the observation that $GL_3(\mathbf F_2)$ the group of automorphisms of $\mathbb F_2^3\cong (\mathbb Z/2 \mathbb Z)^3$ is simple, so the only non injective homomorphism from this group to any other group must be the null homomorphism. We have trivial embedding of $\mathbb F_2^2$ in $\mathbb F_2^3$ and a trivial projection from this to $\mathbb F_2^2$: $$\mathbb F_2^2 \stackrel{i}{\hookrightarrow} \mathbb F_2^3 \stackrel{\pi}{\rightarrow} \mathbb F_2^2 $$

such that $\pi \circ i = 1_{\mathbb F_2^2}$. If there was a functor $\chi$ as above we should get a diagram of type $$\chi(\mathbb F_2^2) \stackrel{\chi{i}}{\rightarrow} \chi(\mathbb F_2^3) \stackrel{\chi{\pi}}{\rightarrow} \chi(\mathbb F_2^2)$$ where $\chi(\pi) \circ \chi (i) = 1_{\chi(\mathbb F_2^2)}$ but $\chi(\pi)$ should be a null homomorphism, because $\chi(\mathbb F_2^3)$ has 168 elements while $\chi(\mathbb F_2^2)$ has just six elements (so the homomorphism cannot be injective). So we get the absurd and it follows that $\chi$ cannot be extended to a functor.

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It is a functor if we restrict the domain category to its subcategory of isomorphisms, however. –  Zhen Lin Jan 5 '13 at 18:44
That's true, but how does it look like when we restrict it to a skeletal category? –  Giorgio Mossa Jan 5 '13 at 18:58
mannaggiaggiorgiomossa –  Ivan Jan 6 '13 at 9:46
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