Re: infinite simple groups have infinite index proper subgroups
- From: "magidin@xxxxxxxxxxxxxxxxx" <magidin@xxxxxxxxxxxxxxxxx>
- Date: Fri, 30 May 2008 16:18:28 EDT
Suppose G is an infinite simple group, and H is a
proper subgroup of G.
Why must [G : H] be infinite?
Suppose G is any group, and H is a subgroup. The
intersection of all conjugates of H is the largest normal
subgroup of G that is contained in H; this may, of course,
be trivial.
The number of conjugates of H in G is equal to the index
of the normalizer of H in G (prove it); the normalizer of
H in G always contains G. Thus, if [G:H] is finite, then
[G:N_G(H)] will also be finite, and the number of
conjugates will be finite.
If G is any group, H and K are subgroups, and [G:H] and
[G:K] are both finite, then [G:H/\K] is finite (again,
prove it). If H is of finite index in G, then any
conjugate of H is also of finite index in G.
So let G be any infinite group, and H a subgroup of finite
index. Then H has only finitely many conjugates, each
of finite index, so their intersection is of finite
index in G; the intersection is normal. That is:
PROP. If G is a group, and H is of finite index in G, then
H contains a subgroup N that is normal in G, and with
[G:N] finite.
So, if G is simple, and H is proper, what is the
conclusion?
Arturo Magidin, sans .sig
magidin@xxxxxxxxxxxxxx
.
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