| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 891.1 | | CTCADM::ROTH | If you plant ice you'll harvest wind | Mon Jun 27 1988 08:32 | 17 |
| � 1. Prove: If H is a subgroup of finite index in a group G,
� then H contains a subgroup N which is of finite index and normal
� in G.
� (Comment: This problem is not hard if you choose the right candidate
� for N, and the candidate is intuitively clear.)
Am I a total idiot - or is it enough to remark that any subgroup contains
the identity?
� 2. Let a, b be real numbers and 0 < a < b.
� etc...
The hand wave is that the map is ergodic - just work on a unit square
(fundamental domain for the covering surface of a torus), then it should
be pretty easy... imbedding in R3 is sort of a red herring.
- Jim
|
| 891.2 | | HPSTEK::XIA | | Mon Jun 27 1988 11:26 | 14 |
| Re .1
Problem 1:
Yes <e> is indeed a subgroup and is normal in G, but <e> is not
necessary of finite index in G.
Problem 2:
Well, indeed it is a problem of ergodics. I have never studied the
subject; therefore, I feel uncomfortable stating the problem in
such form. The torus may or may not be a red herring (I am just not
sure, but chances are you are right.). When I did it, I remember
reducing the problem to proving the ergodicity (sp?) of certain
fucntions on a circle (Warning: I am sure there are many ways of doing
this problem. Do not spend a week reducing the problem in circle :-).
Eugene
|
| 891.3 | Parse trees :-) | ZFC::DERAMO | For all you do, disk bugs for you. | Fri Jul 08 1988 20:39 | 20 |
| Re .0:
>> (Comment: This problem is not hard if you choose the right candidate
>> for N, and the candidate is intuitively clear.)
This does not mean:
The problem is not hard if you choose the right candidate.
and
The [right] candidate is intuitively clear.
Instead, what it must mean is:
The problem is not hard if:
if you choose the right candidate
and
if the candidate is intuitively clear.
Dan
|
| 891.4 | grovel: v. To humble oneself abjectly; cringe | ZFC::DERAMO | Hello, world\n | Thu Jul 21 1988 20:09 | 13 |
| >> 1. Prove: If H is a subgroup of finite index in a group G,
>> then H contains a subgroup N which is of finite index and normal
>> in G.
>> (Comment: This problem is not hard if you choose the right candidate
>> for N, and the candidate is intuitively clear.)
What about, you post the candidate and we supply the
proof?
Dan
P.S. "finite index" means there are only finitely many
cosets of the subgroup.
|
| 891.5 | The Candidate | HPSTEK::XIA | | Mon Jul 25 1988 12:22 | 7 |
| re .4
Well, you take all the groups of the form g^(-1)Hg (The conjugates)
and intersect all of them. That gives you a group. That is the
candidate (You need a normal group, so you push H all over the
place by doing g^(-1)Hg, so....).
Hope that is some help.
Eugene
|
| 891.6 | | LISP::DERAMO | Daniel V. {AITG,LISP,ZFC}:: D'Eramo | Tue Aug 16 1988 19:25 | 14 |
| The results that
-1 -1
a) for each g in G, the set g Hg = {g hg | h is in H}
is a subgroup (called the conjugate subgroup)
b) the intersection of subgroups is a subgroup
are fairly straightforward to show. Here, showing that
this particular intersection gives a normal subgroup
of G isn't difficult, either.
But I haven't shown yet that it has finite index. :-(
Dan
|
| 891.7 | | HPSTEK::XIA | | Fri Aug 19 1988 14:12 | 7 |
| re .6
Hi Dan,
Glad to hear that someone is still trying. I have long given up
the practice of posting Friday afternoon problem.... Oh well, I
will give another hint. There can be only finitely many distinct
groups of the form g^(-1)Hg.....
Eugene
|
| 891.8 | Solution to problem No. 1 | HPSTEK::XIA | | Sat Apr 22 1989 01:08 | 36 |
| Notation: Let A, B be sets, Let AYB denotes the intesection of A and B.
Lemma 1: Let G be a group and H, K be subgroups of finite indices,
then HYK is also of finite indices.
proof: Let {gi}, 1 <= i <= N such that U(Kgi) = G. (U means union).
Now for each i, if (Kgi)YH != nil, let hi be in (Kgi)YH.
then U(Khi) contains H. Let h be in H. Then h = khi for some
k in K and hi in H. ==> k = (hi)^(-1)h ==> k is in H. ==>
U((KYH)hi) = H. Since H is also of finite index, we have KYH
is also of finite index. Q.E.D.
Lemma 2: Let K be a subgroup of G. Let g be an element
of G. If g is in Hgi for some i, then the congugate group
g^(-1)Hg is equal to gi^(-1)Hgi.
proof: Since g is in Hgi, g = hgi for some h in H. g^(-1)Hg =
(hgi)^(-1)H(hgi) = gi^(-1)(h^(-1)Hh)gi = gi^(-1)Hgi.
Hence, if H is of finite indices, the conjugacy class of
H is finite.
Lemma 3: N = Y(g^(-1)Hg) is normal in g where the intesection ranges over
all G.
proof: Let x, g be an elements of G. Let n be an element of N. Then
by definition, n = xg^(-1)h(xg^(-1))^(-1) for some h in H. ==>
x^(-1)nx = x^(-1)xg^(-1)hgx^(-1)x = g^(-1)hg. Hence, x^(-1)nx
belongs to g^(-1)Hg for all g in G. ==> x^(-1)nx is in N ==>
N is normal.
Now lemma 3 gives a normal group N of G. Obviously since H is of finite
index, g^(-1)Hg also has finite index for all g in G. Lemma 2 says that
there are only finite many elements in the conjugacy class of H.
By Lemma 1 and induction, we conclude that N is also of finite index. Q.E.D
Eugene
|