Proving that ${\rm Aut}(G)=\{{\rm id}\} \implies |G| \in \{1,2\}$. The exercise asks to prove that if $G$ is any group with ${\rm Aut}(G) = \{{\rm id}\}$, then $g^2=1$ for all $g$ in $G$, $G$ is abelian, and if $G$ is finite, we'll have $|G| = 1$ or $2$.
I managed to do everything up to the last part (and so far, we can find good answers on this site), but I don't know how to prove that if $|G|>1$, then $|G|=2$. I just know that $|G|$ is even. Can someone give hints? Thanks.
 A: Given $g\in G$, Since conjugation by $g$ is an automorphism of $G$, we have that for all $h\in G$, $ghg^{-1}=h$ and so $gh=hg$. This shows that $G$ is abelian. Now, since $G$ is abelian, the map $x\mapsto x^{-1}$ is an automorphism of $G$, and by hypothesis, $g^{-1}=g\Rightarrow g^2=1$ for all $g\in G$.
Finally, let $S=\{a,b,c_1,\ldots,c_k\}$ be a set of generators of $G$ with $a\neq b$. Notice that the obtained results implies that every element $x\in G$ can be expressed as a product $x=a^ib^jc_x$ where $i,j\in \{0,1\}$ and $c_x$ is a word only using $c_1,\ldots,c_k$ (each one at most once). Let us write furthermore $x'=a^ib^j$.
Consider the permutation $\sigma:=\begin{pmatrix}e&a&b&ab\\e&b&a&ab\end{pmatrix}$. Note that $\sigma$ is an automorphism of the subgroup $G'=\langle a,b\rangle$. Define now the function $\phi:G\to G$ given by
$$\phi(x)=\phi(x'c_x):=\sigma(x')c_x.$$
The map $\phi$ is a bijection, and given $x,y\in G$ we have:
$$\phi(x\cdot y)=\phi(x'y'c_xc_y)=\sigma(x'y')c_xc_y=\sigma(x')\sigma(y')c_xc_y=\phi(x)\cdot\phi(y)$$
So, $\phi$ is an automorphism of $G$, and by hypothesis, we must have $a=\phi(a)=\sigma(a)=b$. Contradiction.
A: If you've managed to do everything up to the last part, then you know that $G$ is an abelian group for which $g^2=e$ for all $g\in G$. This means that $G$ is an $\mathbb{F}_2$-vector space, so you have lots of automorphisms to work with if $|G|\neq 1,2$.
