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It's actually easier to prove the stronger statement that there are infinitely many double cosets $H\backslash F/K$. First, note that if $F'$ is a subgroup of finite index in $F$, with $H'$ and $K'$ …

This is just to flesh out the details of my comment above. I think we can show that $n_k=1$ or $p=1$. After conjugating (and simplifying notation slightly), your equation easily becomes $p^n=[w,b]$ …

This seems to be true and, like many facts about free groups, can be proved using Stallings' famous folding technique. Here's a sketch of the argument. Think of $F(B)$ as the fundamental group of a b …

The case $n=2$ (originally due to Lyndon) admits a very nice geometric argument: one notes that elements $a,b,c$ with $[a,b]=c^2$ lead to a map from the surface $\Sigma_{-1}$ of Euler characteristic - …

As @YCor says in comments, there is a finitely generated example due to Ol'shanskii, which is essentially a kind of Tarski monster. However, Ol'shanskii's construction is very complicated. For "nicer" …

Stallings' folding algorithm (described by Agol) is probably the best way of doing this, but I thought I'd mention an older algorithm which is also useful, due to Whitehead. Let $\{w_k\}$ be a collect …

This is true, though the proof uses some heavy machinery! Theorem A.1 of Agol--Groves--Manning's appendix to Agol's proof of the Virtual Haken conjecture states: Let $G$ be a hyperbolic group, le …

This calculation was performed by Marshall Hall Jr. Let $N(d,n)$ be the number of subgroups of index $d$ in the free group of rank $n$. Then $N(d,n)=d(d!)^{n-1}-\sum_{i=1}^{d-1}((d-i)!)^{n-1}N(i,n)$ …

The assertion about the number of relations seems to be a theorem of Brunner, proved in the paper Brunner, A. M. Transitivity-systems of certain one-relator groups. Proceedings of the Second Internat …

As Derek Holt says in comments, the answer to your first question is 'yes'. You can argue topologically. There is a rose $R$ corresponding to $X$ with $\pi_1R\cong F$. The subset $A$ defines a conn …

This is not strictly an answer to your question, but I hope that results proving that certain nice groups are not IF are also of interest. You point out that free groups and surface groups are IF. Re …

You should look up the work on 'residual finiteness growth' (aka 'Farb growth') by Khalid Bou-Rabee and his coauthors.

The answer to both questions is 'no'. This was proved by Greenberg for Fuchsian groups. One outline of the proof is as follows. Any finitely generated subgroup $H$ of a surface group $\Gamma$ is qua …

As I said in a comment on your other questions, there is no such word. Indeed, for any $w\in F_n$, there is $u\in F_{n-1}$ so that $w$ survives under the retraction $F_n\to F_{n-1}$ that sends the l …

Many examples can be exhibited using a theorem of Gersten: Theorem (Gersten): Let $G$ be a hyperbolic group of cohomological dimension 2. Every finitely presented subgroup $H$ of $G$ is hyperbolic. Th …