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I have been trying to find a counterexample that all groups (finite and infinite) have an abelian subgroup (besides the trivial subgroup).

Any ideas?

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2 Answers 2

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Every non-abelian group has a non-trivial abelian subgroup: Let $G$ be a nonabelian group and $x\in G$, $x$ not the identity. Then $\langle x\rangle$ is an abelian subgroup of $G$.

EDIT: In case you are curious, there are nonabelian groups such that the only abelian subgroups are the cyclic ones generated by one element. For example, $S_3$ is a nonabelian group such that only the cyclic subgroups are abelian.

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    $\begingroup$ Or at least "less trivial". ;) $\endgroup$ Sep 28, 2015 at 3:44
  • $\begingroup$ @Ben Sheller Could you provide more examples other than S3? $\endgroup$
    – byk7
    Feb 1, 2022 at 0:26
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    $\begingroup$ @byk7 Free groups with at least two generators. $\endgroup$ Feb 1, 2022 at 21:42
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It's impossible, as the subgroup generated by one element is cyclic and a cyclic group is abelian.

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