Let $F\_n$ be the free group on $n\geq 2$ generators and let $H < F\_n$ be a finite index normal subgroup. Let $P\subset F$ be the subset consisting of primitive elements (an element of $F\_n$ is called primitive if it is a member of a free basis). Furthermore, we define $\tilde P\subset H$ to be the subset consisting of those powers $x^m$, $m\in \mathbb{Z}$, of primitive elements $x\in P$ which actually are elements of $H$ and we define $\tilde H < H$ to be the subgroup of $H$ generated by $\tilde P$.


**Edit:** We impose another condition which is not satisfied in the excellent counterexample by Mark Sapir given below:


* Suppose that the first basis element $x\_1$ of $F\_n$ is contained in $H$. (This implies $x\_1\in \tilde H$)


Any suggestions, references or counterexamples to either of the following questions are welcome:


1. Is $\tilde H=H$ ? If this not the case:
2. Is $\tilde H$ of finite index in $H$ ? If not:
3. Is the image of $\tilde H$ in $H^{ab}=H / [H,H]$ a subgroup of finite index?



Generating certain subgroups of free groups via primitive elements
If $x\_1 \in H$ then, since $H \unlhd F\_n$, all conjugates of $x\_1$ in $F\_n$ are in $H$ and hence also in $\overline{H}$. So the normal closure $N$, say, of $x\_1$ in $F\_n$ lies in $\overline{H}$.


Now any element of $H$ can be written as $nh$ with $n \in N$ and $h \in \langle x\_2,\ldots,x\_n \rangle$. But $x\_1 h$ is a primitive element of $F\_n$ with $x\_1 h \in H$, so we also have $x\_1 h \in \overline{H}$ and hence $\overline{H}= H$.


