Author 
Topic: Unique subgroup of a finite group (Read 1052 times) 

Michael Dagg
Senior Riddler
Gender:
Posts: 500


Unique subgroup of a finite group
« on: Jan 4^{th}, 2007, 9:21pm » 
Quote Modify

Suppose H is a normal subgroup of a finite group G such that (H,G:H) = 1. Is H the unique subgroup of G having order H ?

« Last Edit: Jan 4^{th}, 2007, 10:41pm by Michael Dagg » 
IP Logged 
Regards, Michael Dagg



ecoist
Senior Riddler
Gender:
Posts: 405


Re: Unique subgroup of a finite group
« Reply #1 on: May 14^{th}, 2007, 8:44pm » 
Quote Modify

Don't know why this problem has gone so long without a posted solution. Just thought of an approach that differs from my first (numbertheoretic) idea for a solution. What about using the following result? Let H be a subgroup of the finite group G which contains the normalizer N(P) of a Sylow psubgroup P of G. Then H is its own normalizer in G. Pardon me for not posting a solution, but I don't want to spoil things for those for whom group theory is a new and fascinating subject.


IP Logged 



Eigenray
wu::riddles Moderator Uberpuzzler
Gender:
Posts: 1948


Re: Unique subgroup of a finite group
« Reply #2 on: May 14^{th}, 2007, 10:35pm » 
Quote Modify

My first thought was that this is "obvious" if G is solvable (Hall), but I didn't think about the general case very much. But then I saw the word "Sylow" and it just clicked: Let K G with K=H. If p  H, then Sylow psubgroups P,P' of H,K are also Sylow psubgroups of G, so they are conjugate in G. But since H is normal, we must have P' K H. Since this holds for all such p, we have H  K H, hence K=H. What did group theorists do before Sylow?


IP Logged 



ecoist
Senior Riddler
Gender:
Posts: 405


Re: Unique subgroup of a finite group
« Reply #3 on: May 15^{th}, 2007, 3:16pm » 
Quote Modify

As usual, Eigenray's solution is the best, but consider the following equally short solution as well. Let H have order n and let k=[G:H]. Let x be any element of G of order dividing n. In the factor group G/H, x^{k}=1 (mod H) and, since x has order dividing n in G, x^{n}=1 (mod H). Since there exist integers u and v such that nu+kv=1, we have x^{1}=(x^{n})^{u}.(x^{k})^{v})=1 (mod H). Hence x lies in H; whence H is the unique subgroup of order n in G.


IP Logged 



Eigenray
wu::riddles Moderator Uberpuzzler
Gender:
Posts: 1948


Re: Unique subgroup of a finite group
« Reply #4 on: May 16^{th}, 2007, 1:29am » 
Quote Modify

Actually I like yours better. It shows that H = {x  x^{n} = 1}. As a followup: Show that G is a Frobenius group, with Frobenius kernel H, iff x^{n}=1 or x^{k}=1 for all x in G.


IP Logged 



Michael Dagg
Senior Riddler
Gender:
Posts: 500


Re: Unique subgroup of a finite group
« Reply #5 on: May 18^{th}, 2007, 12:22pm » 
Quote Modify

Neat solutions!


IP Logged 
Regards, Michael Dagg



