2
$\begingroup$

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?
$\endgroup$
0

2 Answers 2

4
$\begingroup$

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 $\overline{H}= H$.

$\endgroup$
1
  • $\begingroup$ Thank You, this answers my question. I had similar ideas but I didn't realise that an element like $x_1h$ could be primitive -- bit this is clear to me now. If I am not mistaken, your proof shows that in fact $P\cap H$ generates $H$. $\endgroup$
    – Max Smith
    Jun 23, 2013 at 21:01
6
$\begingroup$

Here is a counterexample. Let $\phi$ be the natural homomorphism from $F_2$ to $A=\mathbb{Z}_n\times \mathbb{Z}_n$ where $n$ is a very large odd number (say, $\ge 665$), $H=Ker(\phi)$. Then the image of every primitive element of $F_2$ is a generator of $A$, so it has exponent $n$. Therefore $\tilde P$ is a subgroup of the normal subgroup generated by all $n$-th powers of elements of $F_2$. By Novikov-Adian this subgroup has infinite index.

$\endgroup$
1
  • $\begingroup$ Thank you for this counterexample. In the situations I am interested in, this counterxample can not occur. I have edited my question. $\endgroup$
    – Max Smith
    Jun 23, 2013 at 9:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.