Cross-posted from MSE: <https://math.stackexchange.com/q/1226622/15624>.


Let $G$ be any finite group, and $S\_{\aleph\_0}$ the group of all bijections $\mathbb{Z}\rightarrow \mathbb{Z}$.


Is $G\times S\_{\aleph\_0}\cong S\_{\aleph\_0}$?



direct product of a finite group with an infinite symmetric group
No. The only proper normal subgroups of the infinite symmetric group is the group of permutations with finite support, and the group of even such permutations. Your group also has the normal subgroup $G.$ For the proof of the statement in the first sentence, see Pete Clark's answer to this: [Sign of infinite permutations?](https://mathoverflow.net/questions/12291/sign-of-infinite-permutations)


