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The answer here is negative. In fact, any non-trivial quotient group of $Sym(X)$ contains a copy of $Sym(X)$. Indeed, by the Baer-Schreier-Ulam Theorem, any normal subgroup $N\ne Sym(X)$ is contained in the subgroup $Sym_<(X)$ of permutations having support of cardinality $<\kappa:=|X|$. Let $q:Sym(X)\to Sym(X)/N$ 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 $\pi\in Sym(\kappa)$ of $\kappa$ define the permutation $\bar\pi\in Sym(X)$ 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 $e:Sym(\kappa)\to Sym(X)$, $e:\pi\mapsto\bar\pi$ is a group homomorphism whose image $e(Sym(\kappa))$ in $Sym(X)$ is disjoint with the subgroup $Sym_{<}(X)\supset N$ and hence the composition $q\circ e:Sym(\kappa)\to Sym(X)/N$ is injective.

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, 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.

The answer here is negative. In fact, any non-trivial quotient group of $S(X)$ contains a copy of $S(X)$. Indeed, by the Baer-Schreier-Ulam Theorem, any normal subgroup $N\ne S(X)$ is contained in the subgroup $S_<(X)$ of permutations having support of cardinality $<\kappa:=|X|$. Let $q:S_X\to S_X/N$ 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 $\pi\in S_\kappa$ of $\kappa$ define the permutation $\bar\pi\in S_X$ 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 $e:S_\kappa\to S_X$, $e:\pi\mapsto\bar\pi$ is a group homomorphism whose image $e(S_\kappa)$ in $S_X$ is disjoint with the subgroup $S_{<}(X)\supset N$ and hence the composition $q\circ e:S_\kappa\to S_X/N$ is injective.

The answer here is negative. In fact, any non-trivial quotient group of $Sym(X)$ contains a copy of $Sym(X)$. Indeed, by the Baer-Schreier-Ulam Theorem, any normal subgroup $N\ne Sym(X)$ is contained in the subgroup $Sym_<(X)$ of permutations having support of cardinality $<\kappa:=|X|$. Let $q:Sym(X)\to Sym(X)/N$ 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 $\pi\in Sym(\kappa)$ of $\kappa$ define the permutation $\bar\pi\in Sym(X)$ 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 $e:Sym(\kappa)\to Sym(X)$, $e:\pi\mapsto\bar\pi$ is a group homomorphism whose image $e(Sym(\kappa))$ in $Sym(X)$ is disjoint with the subgroup $Sym_{<}(X)\supset N$ and hence the composition $q\circ e:Sym(\kappa)\to Sym(X)/N$ is injective.

The answer here is negative. For a surjective homomorphism $h:S(X)\to\mathbb Z$ its kernel $N$ is a normal subgroup of $S(X)$. By the Baer-Schreier-Ulam Theorem, the subgroup $N$ is equal either to the subgroup $Alt(X)$ of even finitely supported permutations or to the subgroup $S_\kappa(X)$ of permutations having support of cardinality $<\kappa$ for some infinite cardinal $\kappa\le|X|$.

Since $X$ is infinite, we can choose a family $\{x_{n,\alpha}\}_{(n,\alpha)\in\omega\times|X|}$ in $X$.

For every permutation $\pi\in S_\omega$ of $\omega$ define the permutation $\bar\pi\in S_X$ such that $\bar\pi(x_{n,\alpha})=x_{(\pi(n),\alpha)}$ for $(n,\alpha)\in\omega\times|X|$ and $\bar\pi(x)=x$ for any $x\in X\setminus\{x_{n,\alpha}:(n,\alpha)\in\omega\times|X|\}$. It is clear that $e:S_\omega\to S_X$, $e:\pi\mapsto\bar\pi$ is a group homomorphism whose image $e(S_\omega)$ in $S_X$ is disjoint with the subgroup $S_{|X|}(X)\supset N$ and hence the composition $h\circ e:S_\omega\to\mathbb Z$ is injective, which is a desirable contradiction.

The answer here is negative. In fact, any non-trivial quotient group of $S(X)$ contains a copy of $S(X)$. Indeed, by the Baer-Schreier-Ulam Theorem, any normal subgroup $N\ne S(X)$ is contained in the subgroup $S_<(X)$ of permutations having support of cardinality $<\kappa:=|X|$. Let $q:S_X\to S_X/N$ 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 $\pi\in S_\kappa$ of $\kappa$ define the permutation $\bar\pi\in S_X$ 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 $e:S_\kappa\to S_X$, $e:\pi\mapsto\bar\pi$ is a group homomorphism whose image $e(S_\kappa)$ in $S_X$ is disjoint with the subgroup $S_{<}(X)\supset N$ and hence the composition $q\circ e:S_\kappa\to S_X/N$ is injective.