Re: Normal subgroups of surface groups?

From: James Jiwi (James_Jiwi_at_hotmail.com)
Date: 10/27/04 

Date: Wed, 27 Oct 2004 17:30:07 +0000 (UTC)

You're right. Here is the restatement of the question:
Show that every finitely generated normal subgroup of a non-abelian
surface group (with or without boundary) is of finite index.

JJ

On Tue, 26 Oct 2004 15:00:08 +0000 (UTC), Danny wrote:
>In article <clll9a$fek$1@news.ks.uiuc.edu>,
> James Jiwi <James_Jiwi@hotmail.com> wrote:
>
>> Let G be a surface group, and H be a non-trivial normal subgroup of
G.
>> How can I prove that H is of finite index in G ?
>> (This is a conjecturally true for any one-relator group)
>
>You can't, because this is not true. For instance, any surface group

>(other than the 2-sphere or projective plane) has a surjection to the

>integers, whose kernel is of infinite index. The simplest of course
>would be the torus, where the obvious Z subgroup is normal, and of
>infinite index.
>
>DR

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