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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 $G\times S_{\aleph_0}\cong S_{\aleph_0}$?

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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?

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  • $\begingroup$ The infinite symmetric group has four normal subgroups, by the Baer-Schreier-Ulam theorem. $\endgroup$
    – Angela Pretorius
    Commented Apr 9, 2015 at 19:23
  • $\begingroup$ @Angela this is rather Schreier-Ulam (Baer extended the result to uncountable sets). Actually, this theorem (for a countable set) was first proved by Onofri (1929) and then rediscovered by Schreier-Ulam. See math.stackexchange.com/a/2645097/35400 $\endgroup$
    – YCor
    Commented Feb 10, 2018 at 22:16

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