For any non-empty set $X$ let $\text{Sym}(X)$ denote the group of bijections $f:X\to X$ with composition.


Is there an infinite set $X$ and a surjective group homomorphism $\pi: \text{Sym}(X)\to \mathbb{Z}$?



Surjective group homomorphism from $\text{Sym}(X)$ onto $\mathbb{Z}$
The answer here is negative. In fact, any non-trivial quotient group of the symmetric group $\mathrm{Sym}(X)$ contains a copy of $\mathrm{Sym}(X)$. Indeed, by the [Baer-Schreier-Ulam Theorem](https://groupprops.subwiki.org/wiki/Baer-Schreier-Ulam_theorem), 
any normal subgroup $N\ne \mathrm{Sym}(X)$ is contained in the subgroup $\mathrm{Sym}\_<(X)$ of permutations having support of cardinality $<\kappa:=|X|$. Let $q:\mathrm{Sym}(X)\to \mathrm{Sym}(X)/N$ be the quotient homomorphism.


Since $X$ is infinite, we can choose a family of pairwise distinct $\{x\_{p}\}\_{p\in \kappa\times\kappa}$ in $X$.


For every permutation $\pi\in \mathrm{Sym}(\kappa)$ of $\kappa$ define the permutation $\bar\pi\in \mathrm{Sym}(X)$ letting $\bar\pi(x\_{\alpha,\beta})=x\_{(\pi(\alpha),\beta)}$ for $(\alpha,\beta)\in \kappa\times \kappa$ and $\bar\pi(x)=x$ for any $x\in X\setminus\{x\_{p}:p\in \kappa^2\}$. It is clear that $e:\mathrm{Sym}(\kappa)\to \mathrm{Sym}(X)$, $e:\pi\mapsto\bar\pi$ is a group homomorphism whose image $e(\mathrm{Sym}(\kappa))$ in $\mathrm{Sym}(X)$ is disjoint with the subgroup $\mathrm{Sym}\_{<}(X)\supset N$ and hence the composition $q\circ e:\mathrm{Sym}(\kappa)\to \mathrm{Sym}(X)/N$ is injective.


