My Math Forum Normal subgroup

 Abstract Algebra Abstract Algebra Math Forum

 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 Linear Mode

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

 Contact - Home - Forums - Cryptocurrency Forum - Top