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Finite group proof

  1. Mar 30, 2009 #1
    1. The problem statement, all variables and given/known data

    Let G be a finite group andd H a subset of G. Prove H is a subgroup of G iff H is closed.

    2. Relevant equations



    3. The attempt at a solution
    Let G be a finite group and H be a subgroup.
    G is a finite group, therefore it is closed, has an inverse and has an identity.
    We want to show H is only a subgroup of G iff H is closed.
    To be a subgroup, H must be closed, contain the identity element of G, and contain the inverse.

    Now I'm stuck.
     
  2. jcsd
  3. Mar 30, 2009 #2
    For H to be a subgroup of G existence of inverse is sufficient condition.Because if [tex]a \epsilon\ G[/tex] then [tex] a^{-1}
    \epsilon G[/tex] due to existence of inverse.But since H is closed [tex]a*a^{-1} \epilson \ H[/tex].Therefore [tex]e \epilson H[/tex] .
    Now, since H is a finite group there must exist an n such that [tex]a^n=e[/tex], otherwise it will be an infinite group.So for any [tex]a\epsilon H[/tex] , the inverse is [tex]a^{n-1}[/tex].
     
    Last edited: Mar 30, 2009
  4. Mar 30, 2009 #3
    Wow, that makes more sense now.
     
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