Re: Finite Group

From: William Elliot (marsh_at_privacy.net)
Date: 03/19/05 

Date: Sat, 19 Mar 2005 00:30:49 -0800

From: Arturo Magidin <magidin@math.berkeley.edu>
Newsgroups: sci.math
Subject: Re: Finite Group

>William Elliot wrote:

>> One observes all elements of G have finite order.
>> Pick g /= e and let H = a maximal subgroup not containing g.
>> H is finite. <g,H> the group generated by g and G, equals G.

> I think he attempted to argue as follows: if <g,H> is not all of G,
> there exists x not in <g,H>. Since in particular x is not in H, then
> <x,H> must properly contain H, hence must contain g by maximality of
> H, and from g in <x,H> we deduce that x in <g,H>, contradicting
> choice of x.

Oh, that H is a maximal subgroup of G not containing g,
does not mean that H is a maximal subgroup.

Let H be a maximal subgroup of Q not containing 1/2.
H = { n/m | n,m coprime, m odd }.
<1/2, H> = { n/2m | n,m coprime, m odd } /= Q.

Indeed, divisible groups like Q have no maximal subgroups. 

-- 
G has finite many subgroups iff G is finite.
----