Is it true that every uncountable locally finite group contains a countably infinite normal subgroup? If not, is there a counter example of an uncountable locally finite group that has no countably infinite normal subgroup?
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2$\begingroup$ There exists a profusion of simple uncountable groups. SO(3) is such a group. $\endgroup$– YCorMay 11, 2022 at 23:42
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3$\begingroup$ Thank you @YCor! I didn't know that $SO(3)$ is locally finite. $\endgroup$– Hussain RashedMay 12, 2022 at 0:06
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5$\begingroup$ I mean you're being very polite, but it's clear that $SO(3)$ is not locally finite, right? For instance, there are elements of infinite order (irrational rotations). Maybe YCor just didn't notice this part of your assumption. $\endgroup$– Ronnie PavlovMay 12, 2022 at 0:20
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2$\begingroup$ @bof Yes that's a counterexample, you could answer the question. $\endgroup$– Derek HoltMay 12, 2022 at 7:56
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2$\begingroup$ Sorry, I missed the "locally finite" assumption. Still there are many locally finite ones. For instance, the group of even finitely supported permutations of a set of infinite cardinal $\alpha$ is a simple, locally finite group of cardinal $\alpha$ for every $\alpha$. (Its simplicity is an immediate consequence of those $A_n$ being simple for large $n$.) [Edit: this standard example was proposed in other comments.] $\endgroup$– YCorMay 12, 2022 at 8:28
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