Subsets of free groups contained in $2$-generated subgroups

Question

$\DeclareMathOperator\rank{rank}$Let $F$ be a non-cyclic free group.

For which finitely generated subgroups $H< F$ such that $H$ is not of finite index in a free factor of $F$ does there exist a two-generated proper subgroup $K=\langle x, y\rangle< F$ such that $H\leq K$?

I'd be happy with a proof that it holds for all such finite subsets (unlikely?), and satisfied by a single counter-example. But a nice description of all these subgroups $H$ would be ideal.

The restriction on $H$ not being of finite index in $F$ is clearly necessary, as if $H$ has finite index in $F$ and is contained in a two-generated subgroup $K=\langle x, y\rangle$ of $F$, then we necessarily have $K=F$ (by for example the Schreier index formula). [For a similar reason, if $F$ contained a maximal subgroup $M$ which is finitely generated and not two generated then this would yield a counter-example. However, maximal subgroups of $F$ are not finitely generated.]

More generally, finite index subgroups of free factors have been removed from the question as these can also give counter-examples. For example, if $F=F'\ast F''$ with $F'$ of rank at least $4$ and $H$ has index $2$ in $F'$, then $H$ is not contained in any two-generated subgroup $K$ of $F$ (as otherwise $F=\langle H, a, F''\rangle=\langle K, a, F''\rangle$ where $F'=H\cup Ha$, and so $F$ could be generated by $2+1+\rank(F'')<\rank(F')+\rank(F'')=\rank(F)$ elements).

Answer 1

I think the subgroup $H=\langle a,b^2,ba^2b^{-1}\rangle$ of $F=\langle a,b\rangle$ provides a counterexample.

To see why, first note that $H$ has infinite index in $F$. Now consider the Stallings core graph $\Delta$ for $H$. Let $x,y,z$ denote the three vertices of $\Delta$. For any subgroup $K$ containing $H$, we get an induced map $\Delta\to\Omega$, where $\Omega$ is the core graph for $K$. If this map is an embedding, then $H$ is a free factor of $K$, and in particular the rank of $K$ is at least 3.

On the other hand, if the map isn't an embedding, then some two of $x,y,z$ must get identified. We can therefore proceed by performing each of these three identifications and then folding. What I got when I performed this computation by hand is illustrated here:

Note that, in each case, I get a finite-sheeted covering space of the rose. Therefore, $K$ has finite index in $F$. In particular, if $K$ has rank 2, then $K$ must be $F$ itself, and the inclusion wasn't proper.

To study these subgroups more systematically, I suggest you look at Puder's paper "Primitive words, free factors and measure preservation" and his other papers that refer to it. In particular, the above computation is an example of his algorithm for computing the primitivity rank of a subgroup.

[I just noticed that Sean Eberhard suggested another counterexample in comments. You can check it in exactly the same way.]

Answer 2

Recall that $F_2$ is $\pi_1(\infty)$, where $\infty =S^1 \vee S^1$ is a bouquet of two circles. Let $H \le F_2$ be a subgroup. Then $H \cong \pi_1(\Gamma, v)$ for some covering space $\Gamma$ of $S^1\vee S^1$. Observe that $\Gamma$ is a graph (generally with multiple edges and self-loops). By orienting and colouring the two edges of $\infty$, we get a corresponding orientation and colouring of the edges of $\Gamma$. Sometimes $\Gamma$ is called a Stallings graph (usually all "infinite hanging trees" are omitted from the drawing). Pick a spanning tree $T$ in $\Gamma$ and let $E$ be the set of all edges not contained in $T$. Then $\Gamma$ is homotopically equivalent to a bouquet of $|E|$ circles, so $H \cong F_{|E|}$.

Now let $\{x,y\}$ be a free basis for $F_2$ and let $H$ be a subgroup such that $\langle x, x^y, x^{y^2} \rangle \le H < F_2$. Let $v_1$ be the basepoint of $\Gamma$. Take two steps along edges labelled by $y$ and let the corresponding vertices by $v_2$ and $v_3$. Since $x, x^y, x^{y^2} \in H$, there are $x$-loops based at each of $v_1, v_2, v_3$. Since $y \notin H$, $v_2 \ne v_1$. Now either $v_1 = v_3$ or $v_1 \ne v_3$, but either way we see that $\Gamma$ has at least $3$ edges not in a spanning tree so $H$ has rank at least $3$.