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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?
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2 Answers 2

up vote 4 down vote accepted

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 $\overline{H}= H$.

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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 $H$. –  Max Smith Jun 23 '13 at 21:01

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.

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Thank you for this counterexample. In the situations I am interested in, this counterxample can not occur. I have edited my question. –  Max Smith Jun 23 '13 at 9:16

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