Re: Finite Group
From: Arturo Magidin (magidin_at_math.berkeley.edu)
Date: 03/18/05
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Date: Fri, 18 Mar 2005 16:35:43 +0000 (UTC)
In article <d1ei9m$kr5$1@charm.magnus.acs.ohio-state.edu>,
Zbigniew Fiedorowicz <fiedorow@hotmail.com> wrote:
>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.
>>
>No.
Indeed. In fact, not just no, but "hell, no."
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.
But in fact, from g in <x,H> we deduce g = ax + h for some h in H and
some integer a, which gives that ax is in <g,H>, but it could be that
x is not in <g,H>.
--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================
Arturo Magidin
magidin@math.berkeley.edu
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