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Euge
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Assuming the axiom of choice, show that a group $G$ has trivial automorphism group if and only if $G$ has less than three elements.
mathbalarka said:First, assume that $G$ is abelian. As $|G|$ is at least $3$, it contains $1$, $x$ and the inverse $x^{-1}$ of $x$.The map $x \mapsto x^{-1}$ is an automorphism and it's trivial if and only if $x^2 = 1$ for all $x \in G$. Assume it is not the case, then $x \mapsto x^{-1}$ is the desired nontrivial auto of $G$.
mathbarlarka said:If, on the other hand, every element of $G$ has order $2$, then $G$ is a vector space over $\Bbb Z/2\Bbb Z$, i.e., $G = \bigoplus \Bbb Z/2\Bbb Z$. Consider the map swapping two of the $\Bbb Z/2\Bbb Z$s sitting inside, which is an automorphism. This is always possible for finite $G$, as there are at least two copies of $\Bbb Z/2\Bbb Z$ sitting inside for all $|G| > 2$. However, if $G$ is infinite, picking two copies of $\Bbb Z/2\Bbb Z$ from infinitely many of the copies requires Zorn's lemma (i.e., Axiom of Choice in disguise). So we have proved that if $G$ is abelian, the statement of the problem holds, conditional on AC.
It does not follow from the abelian property of G that |G|≥3.
Since x is a particular point of G, the assignment x↦x−1 is not defined as a function from G into G.
You can't swap copies in G, but elements in G.
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