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Suppose $G$ is a group and let $S$ be the set of all isomorphisms $f: G \rightarrow G$

Show, if $|S|=1$, that the group G is abelian with elements of order 1 or 2

Now so far I know that since $|S|=1$ there is a canonical map or canonical isomorphism but how do I use this fact concretely?

Thank you in advance

KReiser
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napadia
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    Well you need to start thinking. For example, what excludes an abelian group with elements of order greater than $2$? Try some examples to build some idea of what the structural features might be. Also it would be good to know where you have got to in your study of groups to give some context - what do you know about isomorphisms, normal subgroups and the like? – Mark Bennet Mar 05 '21 at 08:18

1 Answers1

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If $G$ is not abelian, then there are two elements $a,b\in G$ such that $ab\neq ba$. Therefore $x\mapsto bxb^{-1}$ is a non-trivial group isomorphism, contradicting $|S|=1$.

On the other hand $x\mapsto x^{-1}$ is an automorphism as well, now we know $G$ is Abelian. So if $|S|=1$ then this automorphism has to be the identity $x\mapsto x$. Meaning for any $x\in G$ we have $x=x^{-1}$ which means that the order is either $1$ or $2$.

Henno Brandsma
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freakish
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