January 2nd, 2010, 12:35 AM  #1 
Newbie Joined: Jan 2010 Posts: 2 Thanks: 0  Normal subgroup
Suppose H is the only subgroup of G of order m, and G=n is finite. How do you prove that H is normal? I've tried a direct proof by considering ghg^1, but don't see a way. I thought about providing an explicit homomorphism of which H is the kernel, I've tried proving that the left and right cosets partition G the same way, and I've tried proving the contrapositive (suppose H not normal in G and prove that there is another subgoup of order m. I could only think to use the fact that for some ghg^1, it is not in H and so try to replace h and it's inverse with this and its inverse and prove that you still get a group, bit I can't prove closure.). At this point I'm just stuck.

January 2nd, 2010, 05:53 AM  #2 
Senior Member Joined: Jan 2009 From: Japan Posts: 192 Thanks: 0  Re: Normal subgroup
Consider the set of elements gHg^1 for a particular g. Does this form a group? (Check for inverses, identity, and closure). If so, what is the order of the group that it forms?


Tags 
normal, subgroup 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
normal subgroup  tinynerdi  Abstract Algebra  5  March 4th, 2010 04:14 AM 
Normal Subgroup  HairOnABiscuit  Abstract Algebra  1  November 25th, 2009 09:46 AM 
Subgroup/Normal Subgroup/Automorphism Questions  envision  Abstract Algebra  3  October 4th, 2009 10:37 PM 
Subgroup/Normal Subgroup/Factor Group Questions  envision  Abstract Algebra  1  October 4th, 2009 03:24 AM 
normal subgroup  fakie6623  Abstract Algebra  3  October 1st, 2007 04:24 PM 