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Prove that every finite group having more than two elements has a nontrivial automorphism.

Proof: Let $G$ is a group such that $o(G)>2$. Let's consider three following cases:

  1. If $G$ is not abelian then exists $a,b\in G$ such that $ab\neq ba$. We can consider the mapping $T_a:G\to G$ defined by $T_a(g)=aga^{-1}$. Then $T_a(b)=aba^{-1}\neq b$ so $T_a$ is nontrivial automorphism.

  2. If $G$ is abelian and there is element $a\in G$ such that $a^2\neq e$ $\Rightarrow$ $a\neq a^{-1}$. Since $G$ is abelian then inverse mapping $T:G\to G$ defined by $T(g)=g^{-1}$ is nontrivial automorphism since $T(a)=a^{-1}\neq a$.

  3. If $G$ is abelian and every $a\neq e$ with $o(a)=2$ or $a=a^{-1}$. I dont know how to continue reasoning.

Remark: I have met at least two topics with the same approaches but the case when $G$ abelian and every non-identity element has order $2$ is solved by considering vector space over $\mathbb{Z}/2$. I am not familiar with vector spaces and I was not able to comprehend this solution. I would be very grateful if somebody will provide more easier solution with detailed explanation.

RFZ
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  • In the case 3. if every element has order 2 then G is abelian – Conjecture Jan 08 '18 at 09:13
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    I doubt you will be able to solve 3 without the knowledge of vector spaces. The main idea behind this approach is to show that $A\simeq\mathbb{Z}_2\oplus\mathbb{Z}_2\oplus A'$ and thus you have a non-trivial automorphism by swapping first two components. – freakish Jan 08 '18 at 09:22
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    I think there is no good reason to avoid vector spaces over $\mathbb{F}_2$. This is the easiest approch, I think. Other arguments are only this in disguise, or at least very much related. – Dietrich Burde Jan 08 '18 at 09:24
  • @DietrichBurde, thanks a lot for comment! I believe that in this moment i will skip this problem and will return when my knowledge about abstract algebra and especially about vector spaces will be better! – RFZ Jan 08 '18 at 10:19
  • @freakish, thanks a lot for comment! I believe that in this moment i will skip this problem and will return when my knowledge about abstract algebra and especially about vector spaces will be better. P.S. What is $A'$ in your text? – RFZ Jan 08 '18 at 11:31

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In your case 3 you ought to be able to prove by a Lagrange Theorem type argument that $|G|=2^k$ and that there exist elements $a_1,a_2,\dots,a_k$ such that every element of $G$ can be written uniquely as $a_1^{e_1}a_2^{e_2}\dots a_k^{e_k}$ where each $e_j \in\{0,1\}$. When we multiply the elements we reduce the exponents modulo $2$.

To conclude check that $$ a_1^{e_1}a_2^{e_2}\dots a_k^{e_k}\mapsto a_1^{e_1}a_2^{e_1+e_2}a_3^{e_3}\dots a_k^{e_k} $$ is an automorphism.

None of this is difficult, but it is really inspired by the vector space argument you've read.

ancient mathematician
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