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Math Help - Quotient groups

  1. #1
    ux0
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    Quotient groups

    Let G be a finite group with K \lhd G. If (|K|,[G : K])=1, prove that K is the unique subgroup of G having order |K|


    I was thinking about approaching this problem by..

    If H \leq G and |H|=|K|, and I figure out what happens to elements of H in  G/K then I might be able to see something happen?.... Or maybe I'm just approaching it all wrong
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  2. #2
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    Quote Originally Posted by ux0 View Post
    Let G be a finite group with K \lhd G. If (|K|,[G : K])=1, prove that K is the unique subgroup of G having order |K|


    I was thinking about approaching this problem by..

    If H \leq G and |H|=|K|, and I figure out what happens to elements of H in  G/K then I might be able to see something happen?.... Or maybe I'm just approaching it all wrong
    suppose H \leq G with |H|=|K|. we have HK \leq G, because K \lhd G. therefore |G|=n|HK|=n\frac{|K|^2}{|H \cap K|}, for some integer n \geq 1. so |H \cap K|[G:K]=n|K| and hence [G:K] \mid n, because

    \gcd([G:K],|K|)=1. let n=m[G:K], for some integer m \geq 1. then |K| \geq |H \cap K|=m|K| \geq |K|. thus |H \cap K| = |K| and hence H=K, because |H|=|K|.
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