Any finite group $G$ can be embedded into $A_{|G|+2}$ via Cayley's theorem ($G\hookrightarrow S_{|G|}\hookrightarrow A_{|G|+2}$). If $G$ is not assumed to be finite, is it still always possible to embedd it into a simple group?

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    $\begingroup$ I'd guess that every infinite group embeds into a simple group of the same cardinality. It's known at least for countable infinite groups. $\endgroup$
    – YCor
    Aug 13, 2016 at 8:15

1 Answer 1


Yes. Assume $G$ is infinite. Cayley still embeds $G$ into the group $S_G$ of permutations of $G$. This group is no longer simple: there is a normal subgroup, call it $N_G$, consisting of all permutations that fix the complement of a subset of $G$ of cardinality smaller than that of $G$. But by a theorem of Baer, Schreier and Ulam, every normal subgroup of $S_G$, other than $S_G$ itself, is contained in $N_G$. Hence $Q_G := S_G / N_G$ is simple. Moreover the composite map $G \to S_G \to Q_G$ is still an embedding because no non-identity element of $G$ has any fixed points. We have thus embedded $G$ into the simple group $Q_G$.

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    $\begingroup$ This is a theorem of Baer. It generalizes the case of the group of permutations of an infinite countable set, initially due to Onofri (1929) and rediscovered by Schreier and Ulam (1933), see math.stackexchange.com/a/2645097/35400 $\endgroup$
    – YCor
    Feb 10, 2018 at 22:11
  • $\begingroup$ If $G$ is countable, will $Q_{G}$ will countable? $\endgroup$
    – user193319
    Jan 26, 2022 at 16:30
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    $\begingroup$ @user193319 no, $S_G$ has the cardinality of the continuum while $N_G$ is countable, so the quotient has the cardinality of the continuum. $\endgroup$
    – grok
    Nov 24, 2022 at 14:04

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