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?
1 Answer
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 nonidentity element of $G$ has any fixed points. We have thus embedded $G$ into the simple group $Q_G$.

4$\begingroup$ 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 $\endgroup$– YCorFeb 10, 2018 at 22:11

$\begingroup$ If $G$ is countable, will $Q_{G}$ will countable? $\endgroup$ Jan 26, 2022 at 16:30

2$\begingroup$ @user193319 no, $S_G$ has the cardinality of the continuum while $N_G$ is countable, so the quotient has the cardinality of the continuum. $\endgroup$– grokNov 24, 2022 at 14:04