This is a question from a textbook: If G is a group and $|Aut(G)|=1$then prove G’=(e)
I’m completely stuck (for about a whole day ), A proper hint or an answer would be so appreciated.
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See https://math.stackexchange.com/questions/3953362/autg-id-iff-g-leq-2/ – Tuvasbien Jan 23 '21 at 05:38
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A nonabelian group has nontrivial inner automorphisms, so if there are no nontrivial automorphisms, the group must be abelian. In fact, having no nontrivial automorphisms implies the group is abelian of exponent $2$, and if you assume the Axiom of Choice, it must have nontrivial automorphisms unless it is cyclic of order $2$ or $1$. – Arturo Magidin Jan 23 '21 at 05:40
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In particular, $G/Z(G)\cong\operatorname{Inn}(G)=1$, which implies $G$ is abelian, and so $G'=1$. – Groups Jan 23 '21 at 05:42