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Math Help - finite groups, normal subgroups, divisors

  1. #1
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    finite groups, normal subgroups, divisors

    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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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by kathrynmath 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 need to show gh^g-1 is in H.
    We have |G|=nb since n is a divisor.
    Merely recall that for any g_0\in G the inner automorphism i_{g_0}:G\to G:g\mapsto g_0gg_0^{-1} is...well..an automorphism. Thus, since H\leqslant G and |H|=n then {g_0}H g_0^{-1}=i_{g_0}(H)\leqslant G and |g_0 Hg_0^{-1}|=n................so?
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    I'm confused. I've never heard the word automorphism
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by kathrynmath View Post
    I'm confused. I've never heard the word automorphism
    Show that g_0 Hg_0^{-1}\leqslant G and has the same order as H.
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  5. #5
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    Quote Originally Posted by Drexel28 View Post
    Merely recall that for any g_0\in G the inner automorphism i_{g_0}:G\to G:g\mapsto g_0gg_0^{-1} is...well..an automorphism. Thus, since H\leqslant G and |H|=n then {g_0}H g_0^{-1}=i_{g_0}(H)\leqslant G and |g_0 Hg_0^{-1}|=n................so?
    Then H is normal
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