@YCor Yes it is.

See the link above for a copy of this question, previously answered. The question is not open, but this does not seem to be very well-known. The exact minimum (which is roughly quadratic in $n$) is determined in the paper eudml.org/doc/282667 "Brunnian links" by Gartside and Greenwood. This question also appeared recently again at math.stackexchange.com/questions/4845211/…

Good point, that argument doesn't quite work.

One can improve this to $\sim \sqrt N$. Choose $p$ minimal so that $p^2+p+1 > N$ and use the Singer construction of a perfect difference in the cyclic group of order $p^2+p+1$. There cannot be a gap larger than about $N^{1/2}$, by the Erdős--Turan bound.

If I understand correctly, a result of Nielsen states that the set of primitive elements $w$ such that $w$ and $x$ map to the same same element of $\mathbb Z^2$ is just the conjugacy class of $x$. (This is special to the rank-two free group.) It follows that $w = x^2 y x^{-1} y^{-1}$ is not primitive. Here "primitive" means "equivalent to $x$ by an automorphism", or equivalently "part of a free basis".

Thanks! By the way, "not part of a free basis" is just a restatement of "not equivalent to $x$".

Maybe this answer should be posted on the original question too (maybe with an extra word or reference about why $\mathrm{Sym}(\omega)$ has the automatic continuity property?), since it is substantially different from the answer given there.

@YCor Is there a countable quotient of the direct product of all finite groups that covers infinitely many alternating groups?

I agree with @YCor that my answer should not be accepted in its current form. Although I think it almost certain that such a group $S$ does not exist, it's not obvious, and I would be interested to see the demonstration. Also, I am actually not at all sure that $G = \prod_F F$ has no countable quotient covering every finite group, it was just a thought.

For the number of abelian groups, see math.stackexchange.com/q/119642/23805.