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Finite groups

  1. Jan 6, 2010 #1
    I've been trying to prove something that seems obvious but have had no success thus far:

    say G is a finite group and H and K are proper subgroups, if K contains a conjugate of H, then it isn't possible to have G=HK.

    Proof anybody? I'm happy if one can prove the special case below:

    It's fairly easy to show one can't have the product of a proper subgroup and it's conjugate equal to a group i.e. it isn't possible that Hx^{-1}Hx=G unless H=G. Can you show it for an arbitrary product of conjugates? i.e. if [tex]\Pi_{x \in G} H^x=G[/tex] then H=G. Note that [tex]H^x=x^{-1}Hx[/tex]

    thanks for the help
  2. jcsd
  3. Jan 6, 2010 #2
    Let c be an element such that [itex]cHc^{-1} \subseteq K[/itex].

    Assume G=HK.

    Express [itex]c^{-1}[/itex] as a product [itex]c^{-1}=hk[/itex] with h in H and k in K. Then
    [tex]1 = chc^{-1}ck[/tex]
    [tex](chc^{-1})^{-1}k^{-1} = c[/tex]
    both factors on the left hand side are in K so c is in K.

    Since c is in K, [itex]H \subseteq c^{-1}Kc = K[/itex]. We then have G=HK=K so K isn't a proper subgroup of G.
  4. Jan 7, 2010 #3
    thanks for the prompt reply, i appreciate the help.
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