# Every finite group with more than $2$ elements has a non-trivial automorphism.

$$\newcommand{\Aut}{\operatorname{Aut}}$$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 $$G$$.

• Another instance of this question being asked before might be helpful: math.stackexchange.com/questions/395569/… Commented Aug 27, 2021 at 5:16
• @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.
– Koro
Commented Aug 27, 2021 at 5:30

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 $$x=h^{-1}a=ha$$.