Universal group such that every finite group is a quotient

Question

We say that a permutation $\varphi:\mathbb{N}\to\mathbb{N}$ is finitary if there is $k\in\mathbb{N}$ such that $\varphi(i) = i$ for all $i\in\mathbb{N}$ with $i\geq k$. Let $I_\mathbb{N}$ denote the group of finitary permutations of $\mathbb{N}$, with composition as group operation. Every finite group can be embedded into $I_\mathbb{N}$.

Turning arrows around, is there a group $S_\mathbb{N}$ with the following strong properties?

1. For every finite group $F$ there is a surjective group homomorphism $\pi:S_\mathbb{N}\to F$, and
2. If $G^*$ is a group such that for every finite group $F$ there is a surjective group homomorphism $\pi:G^*\to F$, then there is a surjective group homomorphism $s:G^*\to S_\mathbb{N}$.

(Note that in the embedding case, there is no such group, see the comment section; in particular $I_\mathbb{N}$ is not "universal" in the above sense.)

Answer 1

Not a complete answer: Let $G = \prod_F F$ be the direct product of all finite groups and let $F_\omega$ be the free group on a countably infinite collection of generators. Obviously $G$ and $F_\omega$ both cover every finite group. To show that there is no group $S$ as sought in the question it suffices to show that $G$ has no countable quotient that still covers every finite group. Does it?

Answer 2

Again not a complete answer: If we consider the analog situation, where we replace the finite groups by simple finite groups, then there exists no such universal group $S$.

Indeed, assume such a hypothetical universal group $S$ exists. Then it must be countable by an analogous argument to the answer of Sean Eberhard and of course $S$ cannot be finite. Analogously, as above, we consider $G = \prod_{F \in \mathscr{F}} F$, where the product runs over a system of representatives $\mathscr{F}$ of all finite simple groups up to isomorphism. We call $S \subseteq G$ a product subgroup, if $S = \prod_{F \in \mathscr{F}} S_F$ where $S_F$ is a subgroup of $F$. A product subgroup $S$ of $G$ is normal if and only if $S_F$ is trivial or equal to $F$ for all $F \in \mathscr{F}$. In the next paragraph we show that every normal subgroup of $G$ is a product subgroup of $G$. This will prove that every quotient of $G$ is either uncountable or finite, and hence, there exists no surjective homomorphism $G \to S$, contradiction.

Let $N$ be a normal subgroup of $G$. Choose a maximal normal product subgroup $S$ of $G$ such that $S \cap N$ is trivial. Consider the quotient map $\pi \colon G \to G/S$. Then $\pi$ restricts to an injection $N \to G/S$ which is also surjective by the maximality of $S$ (in fact, $G/S = \prod_{F \in \mathscr{F}_S} F$ for some subset $\mathscr{F}_S$ of $\mathscr{F}$. If $N \to G/S$ is not surjective, then there exists $F_0 \in \mathscr{F}_S$ such that $T = \prod_{F \in \mathscr{F}_S} R_F$ intersects $\pi(N)$ trivially, where $R_F$ is trivial for all $F \in \mathscr{F}_S \setminus \{F_0\}$ and $R_{F_0} = F_0$. But in this case $\pi^{-1}(T)$ is a normal product subgroup of $G$ that intersects $N$ trivially but contains $S$ properly, contradiction). Let $\rho \colon G \to \prod_{F \in \mathscr{F} \setminus \mathscr{F}_S} F$ be the natural projection. Then the homomorphism $$ \prod_{F \in \mathscr{F}_S} F \simeq N \xrightarrow{\rho|_N} \prod_{F \in \mathscr{F} \setminus \mathscr{F}_S} F $$ is trivial, since every homomorphism $F \to F'$ is trivial for distinct $F, F' \in \mathscr{F}$. This shows that $N$ lies in the kernel of $\rho$. Since, $N$ and $\ker(\rho)$ are both sections of $\pi \colon G \to G/S$, we get $N = \ker(\rho)$. Hence, $N$ is a normal product subgroup of $G$.

EDIT: The argument doesn't work, as the homomorphism $\rho |_N \colon N \to \prod_{F \in \mathscr{F} \setminus \mathscr{F}_S} F$ is not necessarily trivial!