# Can any group be embedded in a simple group?

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?

• 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.
– YCor
Aug 13, 2016 at 8:15

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 $Q_G$.
• If $G$ is countable, will $Q_{G}$ will countable? Jan 26, 2022 at 16:30
• @user193319 no, $S_G$ has the cardinality of the continuum while $N_G$ is countable, so the quotient has the cardinality of the continuum.