March 30th, 2007, 12:03 AM  #1 
Newbie Joined: Feb 2007 Posts: 8 Thanks: 0  Finitely Generated
Suppose G is finitley generated group for every n prove that there are finite subgroups like H such that [G:H]=n

March 30th, 2007, 05:11 AM  #2  
Member Joined: Dec 2006 Posts: 39 Thanks: 0  Re: Finitely Generated Quote:
Now, it is NOT true that if G is a f.g. group then for any n (natural number?) there's a sbgp. of index n. Just take some finite group and let n be a natural number that does NOT divide the order of G... Tonio  
March 30th, 2007, 06:58 AM  #3 
Site Founder Joined: Nov 2006 From: France Posts: 824 Thanks: 7 
I tried to add the infiniteness but I am unable even to start. If that kind of problem is not open, there must be some pretty heavy results behind ...

March 30th, 2007, 11:20 AM  #4 
Newbie Joined: Feb 2007 Posts: 8 Thanks: 0 
I mean G is a group which is finitely generated and n is natural number .prove that there are finite subgroups like H such that [G:H]=n. 
March 31st, 2007, 10:54 AM  #5  
Member Joined: Dec 2006 Posts: 39 Thanks: 0  Quote:
Again, this is NOT true. If G is finite, then there are plenty of counterexamples for plenty of natural numbers....for example, no group of order 36 has a sbgp. of index 7,5, 11 or 23. Now if G is infinite, then if H is finite the index [G:H] HAS to be infinite, so the question AGAIN makes no sense Tonio  
July 22nd, 2007, 02:02 AM  #6 
Newbie Joined: Jul 2007 Posts: 1 Thanks: 0 
let G=<a1,...,am>. If G:H=n, we can get the generators of H by the coset representatives. We know the are finite many different representatives by ReidemesterSchreier system.

August 13th, 2007, 02:56 AM  #7  
Member Joined: Dec 2006 Posts: 39 Thanks: 0  Quote:
Revisiting the site after sometime I think I may, probably, know what the OP meant to ask: he wants to prove that if G is a f.g. group, then for any natural number n there's a finite number of sbgps. H of G s.t. [G:H] = n. If this is what he meant to ask....wow, dude! Please do improve your english! Anyway you'll need it to make a reasonably worthwhile career in maths since a huge ammount of professional papers and books are in english. Anyway: Let H <= G be s.t. [G:H] = n ==> by the regular representation of G in the set of left cosets of H in G, we get a homomorphism G > S_n. Since any homom. from G to any group is uniquely determined by its action on any set of generators of G, and since G is fin. generated (say, by g1,..., gr), there thus are at most (n!)^r homom. from G into S_n, and since any sbgp. H as above gives rise to one of these, there are at most (n!)^r sbgps. of G of index n. Regards Tonio  
September 19th, 2007, 05:21 AM  #8 
Newbie Joined: Sep 2007 Posts: 3 Thanks: 0 
Tonio, I can't see how the representation you mention uniquely determines H. There might be more than one H giving rise to the same permutation representation. This is true already for the subgroups of order 2 in Z/2Z x Z/2Z. /d 

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