# Every finite group of order more than two has a nontrivial automorphism [duplicate]

This question already has an answer here:

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.

## marked as duplicate by user61527, Ayman Hourieh, Camilo Arosemena-Serrato, Lost1, Moishe KohanFeb 16 '14 at 3:29

• This has been asked more than once before math.stackexchange.com/questions/391453/… – David Peterson Feb 16 '14 at 1:10
• 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 1:38
• 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 9:58
• 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 5:55

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 $\mathbb Z/2\mathbb Z$.
• $\mathbb{Z}$-modules are not yet covered at this point in the text. – AG. Feb 16 '14 at 1:37