Let G be a finite group and let n be a divisor of |G|. Show that if H is the only subgroup of G of order n, then H must be normal in G. I need to show gh^g-1 is in H. We have |G|=nb since n is a divisor.
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Originally Posted by kathrynmath Let G be a finite group and let n be a divisor of |G|. Show that if H is the only subgroup of G of order n, then H must be normal in G. I need to show gh^g-1 is in H. We have |G|=nb since n is a divisor. Merely recall that for any the inner automorphism is...well..an automorphism. Thus, since and then and ................so?
I'm confused. I've never heard the word automorphism
Originally Posted by kathrynmath I'm confused. I've never heard the word automorphism Show that and has the same order as .
Originally Posted by Drexel28 Merely recall that for any the inner automorphism is...well..an automorphism. Thus, since and then and ................so? Then H is normal
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