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Well, this is an exercise problem from Herstein which sounds difficult:

  • How does one prove that if $|G|>2$, then $G$ has non-trivial automorphism?

The only thing, i know which connects a group with its automorphism is the theorem, $$G/Z(G) \cong \mathcal{I}(G)$$ where $\mathcal{I}(G)$ denotes the Inner- Automorphism of $G$. So for a group with $Z(G)=(e)$, we can conclude that it has a non-trivial automorphism, but what about groups with center?

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You've got it switched. The group of inner automorphisms is isomorphic to G/Z(G). This essentially solves the problem. –  Qiaochu Yuan Oct 30 '10 at 19:21
    
@Qiaochu Yuan: Sorry –  anonymous Oct 30 '10 at 19:23
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So figure it out. Inner automorphisms are automorphisms, so if G/Z(G) is nontrivial there is an inner automorphism. Otherwise... –  Qiaochu Yuan Oct 30 '10 at 19:25
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@Chandru1: So now you have reduced to the case when $G$ is abelian. Can you think of an automorphism that works for abelian groups, but not for nonabelian groups? –  Jonas Meyer Oct 30 '10 at 19:27
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@Chandru1: Please think more about where it is reduced to the abelian case. Yes, that is the map I had in mind. –  Jonas Meyer Oct 30 '10 at 19:29

2 Answers 2

up vote 20 down vote accepted

As you note in the question, the group of inner automorphisms Inn($G$) is isomorphic to $G/Z(G)$. In particular, it's trivial if and only if $Z(G)=G$. So there is a non-trivial (inner) automorphism unless $G=Z(G)$.

Now, notice that, by definition, $Z(G)=G$ if and only if $G$ is abelian; so we have reduced to the abelian case.

If $G$ is abelian then $g\mapsto -g$ is an automorphism, and it is non-trivial unless $g=-g$ for all $g\in G$. But $g=-g$ if and only if the order of $g$ divdes two. So we have now reduced to the case in which $2g=0$ for all $g\in G$.

In this case, $G$ is a vector space over the field $\mathbb{Z}/2$. As $|G|$ is equal to 2 raised to the power of the $\mathbb{Z}/2$-dimension of $G$, the hypothesis that $|G|>2$ implies that $\mathrm{dim}_{\mathbb{Z/2}} G>1$. But now we can write down lots of linear automorphisms of $G$. For instance, you could fix any basis $g_1,g_2,\ldots$ and take the automorphism $g_1\mapsto g_2$, $g_2\mapsto g_1$ and $g_i\mapsto g_i$ for every $i>2$.

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HJRW, I'm sorry, but I couldn't grasp the details of your reasoning when $G$ is abelian and $g^2 = e$ for all $g\in G$. I've never encountered this vector space business getting into group theory. So can you please explain your reasoning to me in more elementary terms? –  Saaqib Mahmuud Jun 14 at 13:59
    
@SaaqibMahmuud; well, suppose that $G$ is abelian and $2g=0$ for all $g$. Then you can check very easily that your group $G$ satisfies the axioms of a vector space over the field with two elements (which I denoted by $\mathbb{Z}/2$ in my answer). Then you argue exactly as I described above. If there are any further steps you have difficulty with, it would help if you said which ones they are! –  HJRW Jun 15 at 5:46
    
Much easier to use the fundamental theorem of abelian groups for exponent 2. Then you obviously have permutation isomorphisms for order greater than 2. Same idea in a sense, but you don't have to leave group theory. –  zibadawa timmy Aug 25 at 5:40

If $G$ is not abelian, then conjugation by a noncentral element will do.

If $G$ is abelian, then $x\mapsto x^{-1}$ is an automorphism. It will be nontrivial unless every element of $G$ equals its inverse, that is, if every element of $G$ is of exponent $2$.

If every element of $G$ is of exponent $2$, then $G$ is a vector space over the field of $2$ elements, so it is isomorphic to a (possibly infinite) sum of copies of $C_2$, the cyclic group of two elements. Since $|G|\gt 2$, there are at least two copies, so the linear transformation that swaps two copies of $C_2$ is a nontrivial automorphism.

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Beat me to it by 90 seconds! –  HJRW Oct 30 '10 at 21:17
    
@Henry: Looks that way! (-: –  Arturo Magidin Oct 30 '10 at 22:20
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Any idea whether this can be proven constructively? –  Carl Mummert Oct 30 '10 at 23:49
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Carl, what do you mean? There are three cases, but in each case the automorphism is completely explicit. The existence of complete groups, in which every automorphism is inner, shows that there is no one construction that can work for all groups (as abelian groups have no inner automorphisms). –  HJRW Oct 31 '10 at 1:05
    
Henry Wilton: I mean, e.g., in Bishop's system. But one could also ask the analogous computability question whether there is a computable procedure that assigns to every index of a computable group of more than two elements an index of a nontrivial automorphism. While it seems like there shouldn't be a procedure like that, I don't quite see how to prove it. I wanted to ask Arturo first before asking it as a separate question. –  Carl Mummert Oct 31 '10 at 1:26

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