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I want to prove that every finite group $G$ of order more than 2 has a nontrivial automorphism. I've seen this question answered on this site for infinite groups, but the proofs given use the fact that if $g^2=1$ for every $g$ in $G$, then $G$ is a vector space over $\mathbb{Z}_2$. This is an exercise in Herstein's text that appears before the section on (the fundamental theorem of) finite abelian groups. I think I can prove this result using that theorem, but was wondering if there are more elementary proofs.

Here is my proof: If $G$ is nonabelian, then $\exists x \in G$ such that the map $(T_x: g \mapsto x^{-1}gx)$ is a nontrivial automorphism of $G$. So suppose $G$ is abelian. Then the map $g \mapsto g^{-1}$ is an automorphism of $G$; this automorphism is nontrivial if some element in $G$ has order at least 3. If every element in $G$ has order 2, then by the fundamental theorem on finite abelian groups, $G \cong C_2 \times \cdots \times C_2$ is the direct product of $k$ copies of $C_2$ for some $k \ge 2$. A map that interchanges the generators of the first two copies and fixes the remaining $k-2$ copies yields a nontrivial automorphism of $G$. QED.

AG.
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    This has been asked more than once before http://math.stackexchange.com/questions/391453/non-trivial-automorphism – David P Feb 16 '14 at 01:10
  • I think my question is not a duplicate because I am asking for a proof that does not appeal to the theorem on vector spaces over $\mathbb{Z}/2$ or the fundamental theorem on finite abelian groups (which are used in the proofs given on this website so far). – AG. Feb 16 '14 at 01:38
  • The first paragraph of my question makes it clear I would like to see a proof (for finite groups) which does not use the fundamental theorem of finite abelian groups. So the proof referenced in the other webpage can't be used to answer my question. – AG. Feb 16 '14 at 09:58
  • The accepted answer on the other page does not use the fundamental theorem of finite abelian groups. – Jim Feb 16 '14 at 18:17
  • @Jim The accepted answer on the other page does use the fact that if $G$ is abelian and has exponent 2, then $G$ is a vector space over $\mathbb{Z}/2$. This fact is similar to the fundamental theorem of finite abelian groups. – AG. Feb 17 '14 at 05:55
  • The fact you're talking about is fairly trivial and it's "proof" is just routine checking that axioms hold. I would hardly call that similar to a structure theorem for all finite abelian groups, but I guess it's a matter of opinion. – Jim Feb 17 '14 at 06:58

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Your proof is correct, and with a slight adjustment applied to both finite and infinite groups: In the last step you don't need the fundamental theorem of finite abelian groups. An abelian group is a $\mathbb Z$-module. If every element has order $2$ then it's a $\mathbb Z/2\mathbb Z$-module, so a vector space, and the only vector space infinite or otherwise that doesn't have a nontrivial automorphism is $\mathbb Z/2\mathbb Z$.

Jim
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  • Why do you mention infinite groups? I see what you mean, later, but the problem is that, up front, it seems a little odd to bring it up - like a non sequitur. – Thomas Andrews Feb 16 '14 at 01:29
  • $\mathbb{Z}$-modules are not yet covered at this point in the text. – AG. Feb 16 '14 at 01:37
  • @ThomasAndrews: If you read the post the OP says he's seen this question answered for infinite groups. – Jim Feb 16 '14 at 01:57
  • Ah, yes, I missed that section, but you just repeat the proof that he references there, too. – Thomas Andrews Feb 16 '14 at 02:05
  • lol, yea, I missed that section – Jim Feb 16 '14 at 07:36
  • The first paragraph of my question makes it clear I would like to see a proof (for finite groups) which does not use the fundamental theorem of finite abelian groups. So the proof referenced in the other webpage can't be used to answer my question. – AG. Feb 16 '14 at 09:57