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Math Help - [SOLVED] Prove a finite subgroup is normal, given order conditions.

  1. #1
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    [SOLVED] Prove a finite subgroup is normal, given order conditions.

    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'm not really sure what to do with this. It's been a while since I worked with cyclic groups or Lagrange's theorem, and so I may be forgetting some essential tricks.

    Any help would be much appreciated!
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by hatsoff View Post
    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'm not really sure what to do with this. It's been a while since I worked with cyclic groups or Lagrange's theorem, and so I may be forgetting some essential tricks.

    Any help would be much appreciated!
    Let i_g\left[H\right] denote the inner automorphism of H. In other words, i_g\left[H\right]=gHg^{-1}. First prove that i_g\left[H\right] is a subgroup of G and then observe that the mapping \phi:H\mapsto  i_g\left[H\right] is a bijection. In other words we have that gHg^{-1} is a subgroup with the same order as H. Thus, by hypothesis we have that gHg^{-1}=H and the conclusion follows.
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    Thanks man! I guess I just have to better familiarize myself with this stuff.
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