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?

2$\begingroup$ 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 '16 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 nonidentity element of $G$ has any fixed points. We have thus embedded $G$ into the simple group $Q_G$.

1$\begingroup$ 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 '18 at 22:11