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Groups of order 21 (Need help understanding an inference)

  1. Aug 18, 2011 #1
    Sylow's theorem tells us that there is one 7-Sylow subgroup and either one of seven 3-Sylow subgroups. Call these subgroups H and K respectively. Sylow's theorem also tells that H is normal in G.

    I'm not going to write it all out as I don't think it's necessary but in the case when we have seven 3-Sylow subgroups, we conclude that the generators [itex]x[/itex] (of order 7) and [itex]y[/itex] (of order 3) generate the entire group G. Since H is normal, we have know that [itex]yxy^{-1} = x^k[/itex], for some [itex]k[/itex], [itex]0 \leq k \leq 6[/itex]. [itex]k[/itex] cannot equal 0 and 1 because that would imply [itex]x = e[/itex] in the first case and G is abelian in the second case, contrary to assumption in both cases.

    Here's where I get lost:

    Since [itex]y[/itex] has order 3, and [itex]y^3 x y^{-3} = x^{k^3}[/itex] ...

    How did he infer that [itex]y^3 x y^{-3} = x^{k^3}[/itex] from what was given?
     
  2. jcsd
  3. Aug 18, 2011 #2

    micromass

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    Well, first notice that for every n, it holds that

    [tex](yxy^{-1})^n=yx^ny^{-1}[/tex]

    thus

    [tex]
    \begin{eqnarray*}
    x^{k^3}
    & = & ((x^k)^k)^k\\
    & = & ((yxy^{-1})^k)^k\\
    & = & y(x^k)^ky^{-1}\\
    & = & y(yxy^{-1})^k y^{-1}\\
    & = & y^2x^ky^{-2}\\
    & = & y^3 x y^{-3}
    \end{eqnarray*}
    [/tex]

    Does that help??
     
  4. Aug 18, 2011 #3
    Oh doh! I forgot about that property. So to answer your question, yes it does help! Thank you very much!
     
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