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January 2nd, 2010, 12:35 AM   #1
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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.
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January 2nd, 2010, 05:53 AM   #2
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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?
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