Direct product of a finite group with an infinite symmetric group

Question

Let $G$ be any finite group, and $S_{\aleph_0}$ the group of all bijections $\mathbb{Z}\rightarrow \mathbb{Z}$.

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

Answer 1

Question has been answered at MO:

$\ \ S_{\aleph_0}$ has only the four normal subgroups given by the Baer-Schreier-Ulam theorem whereas $G\times S_{\aleph_0}$ has the normal subgroup $G$.

https://mathoverflow.net/questions/202472/direct-product-of-a-finite-group-with-an-infinite-symmetric-group.

Answer 2

If $X$ is any set (with no exception, possibly uncountable, possibly finite, even with less than 4 elements), then the set of normal subgroups of $S_X$ is totally ordered by inclusion. On the other hand, if $F$ is any group (finite or not) and if both $F$ and $X$ have at least two elements, then the set of normal subgroups of $S_X\times F$ is not totally ordered, since it contains both $S_X$ and $F$, none of which being contained in the other one; more generally this shows that $S_X$ is not isomorphic to the direct product of any two nontrivial groups.