Suppose $G$ is a group $w \in F_\infty$, where $F_\infty$ is the free group of countable rank. Let’s define the corresponding word map as $w(G) := \{g \in G| \exists f \in Hom(F_\infty, G) f(w) = g\}$, where $Hom(H, K)$ is the set of all homomorphism from $H$ to $K$.
Now, let’s for arbitrary cardinality $\alpha \geq \aleph_0$ define $S_\alpha$ as the group of all permutations of a set of cardinality $\alpha$.
According to a theorem proved by R. Baer in «Die Kompositionsreihe der Gruppe aller einendeutigen Abbildungen einer unendlichen Menge auf sich», the unique largest proper normal subgroup of $S_\alpha$ is the subgroup $S_{<\alpha}$ of all permutations with cardinalities strictly less than $\alpha$. It is also not hard to see, that $\frac{S_\alpha}{S_{<\alpha}}$ contains a subgroup isomorphic to $S_\alpha$ and $S_\alpha$ contains subgroups isomorphic to all groups of order $\alpha$ or less by Cayley theorem. Thus $\frac{S_\alpha}{S_{<\alpha}}$ generates the variety of all groups. That means that all non-trivial verbal subgroups of $S_\alpha$ are equal to the whole group. Thus for any non-trivial group word $w$, $\langle w(S_\alpha) \rangle = S_\alpha$.
However, for the combatant a stronger statement is true. Oystein Ore proved in «Some remarks on commutators», that $[x, y](S_\alpha) = S_\alpha$.
My question is:
Is it true, that $\forall w \in F_\infty \setminus \{e\}$ $w(S_\alpha) = S_\alpha$?
No. For instance, any square in $S_\alpha$ must have an even number of infinite cycles, so not every element of $S_\alpha$ can be represented by the word $w=a^2$ (or similarly, $w=a^n$ for any $n>1$).