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I'm having problems with the first implication. I already know how to prove that $|G| \leq 2$ implies $Aut(G) = \{id\}$. But how can I prove the converse? I found it kind of simple but also a little hard to write. I did this:

Because $Aut(G) = \{id\}$ then there is only one way to rearrange all the elements, and by definition of $Aut(G)$, then $|G| \leq 2$, because you have to fix the neuter and the other element should be fixed if $|G| = 2$.

But I'm not very convinced myself. Anny help would be appreciated.

RobPratt
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Albedo
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2 Answers2

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Note that $G/Z(G) \cong Inn(G)\leq Aut(G) = \{id\}$, so $G$ must be abelian. In an abelian group, you always have the autmorphism sending each element to its inverse. In order for this to be the trivial automorphism, each element must equal its inverse, so $G$ has only elements of order $2$, i.e. it is isomorphic to some number of copies of $\mathbb{Z}/2\mathbb{Z}$. If there is more than one copy, you can construct a nontrivial isomorphism which swaps copies of $\mathbb{Z}/2\mathbb{Z}$, which leaves two possibilities, either $G \cong \mathbb{Z}/2\mathbb{Z}$ or $G \cong \{id\}$.

Noah Solomon
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Let $x\in G$, then $y\mapsto xyx^{-1}$ is an automorphism of $G$ and thus is trivial. Therefore for all $y\in G$, $xyx^{-1}=y$ and $G$ is commutative. We can finally consider the automorphism $x\mapsto -x$, it is trivial and therefore $2x=0$ for all $x\in G$ and $G$ is a $\mathbb{Z}/2\mathbb{Z}$-vector field, thus there exists $I$ such that $G\simeq(\mathbb{Z}/2\mathbb{Z})^I$. If ${\rm Card}(I)\geqslant 2$, then ${\rm Aut}(G)\neq\{{\rm Id}\}$ because we can swap two distinct elements of $G$ and this gives a non-trivial automorphism, therefore ${\rm Card}(I)\leqslant 1$ and $o(G)\leqslant 2$.

Tuvasbien
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