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$\newcommand{\Aut}{\operatorname{Aut}}$Theorem: Given a finite group $G$ of order $n\gt 2$, it follows that $\left|\Aut G \right| \gt 1$.

Proof: I tried it like this:

Suppose on the contrary that $\left|\Aut G\right|\le1$. It follows that $\left|\Aut G\right|=1$ as identity permutation $i:G\to G$ defined as $i(x)=x$ , always belongs to $\Aut G$.

I know the result that: $G/Z(G)\sim I(G)$, where $I(G)$ is the set of all inner automorphisms of $G$; and that $I(G)\le \Aut G$. It follows that $G=Z(G)\implies G$ is abelian.

It can be shown here that $T:G\to G$ defined by $T(x)=x^{-1}$ is an automorphism. But as per the assumption, it should follow that $T(x)=x$ for all $x\in G$, that is, $x=x^{-1}$ for all $x\in G\implies$ every non-identity element of $G$ is of order $2.\implies G$ is of even order.

Above shows that if $G$ (finite and of order greater than $2$) is a non-abelian group or a group of odd order, then the theorem is proved by contradiction.

So all it remains is to prove the result when $G$ is abelian and every non-identity element is of order $2$.

$G=\{1,a_1,a_2,\dots,a_{2k-1}\}$, where $k\ge 2$ and $2k=n$. I tried to create a non-trivial automorphism as follows:

Let $T:G\to G$ be defined as $T(x)=\begin{cases} x \text{ if $x\notin \{a_1,a_2\}$}\\a_2 \text{ if $x=a_1$}\\a_1 \text{ if $x=a_2$}\end{cases}$

$T$ is clearly a bijection. I tried to prove it homomorphism but got stuck and I feel that it's not a homomorphism (as it's not yet clear where $xy$ will be mapped by $T$ when $x$ and $y$ both are neither $a_1$ nor $a_2$). How can I create a not trivial automorphism of $G$? Any suggestions are welcome. Thanks.

I have seen this question being asked before (here) but the answers use arguments on finite field which I don't currently know much about. Therefore, I tried to construct an explicit non-trivial automorphism of $G$.

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  • $\begingroup$ Another instance of this question being asked before might be helpful: math.stackexchange.com/questions/395569/… $\endgroup$
    – TomKern
    Aug 27, 2021 at 5:16
  • $\begingroup$ @TomKern: Thanks for the link. I think what I am missing is that $G$ should be isomorphic to an external product of $Z_2$'s, whose proof can be given using "fundamental theorem for abelian groups" which I have not covered yet. $\endgroup$
    – Koro
    Aug 27, 2021 at 5:30

1 Answer 1

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The proof of "the fundamental theorem for abelian groups" for groups each element of which has order two is very simple.

So, let $G$ be such a group. Then it is abelian. Let $a\in G$, $a\neq1$ and $H$ be the maximal subgroup of the group $G$ with property $a\notin H$. Let us prove that $G=H\cdot\langle a\rangle$. If $x\in G$, then due to the maximality of $H$, we have $a\in \langle H,x\rangle$. It follows that $a=hx$, $h\in H$ and $x=h^{-1}a=ha$.

Next, induction by the order of the group works.

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