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Sylow's Theorem proof

  1. Apr 18, 2012 #1
    1. The problem statement, all variables and given/known data

    Show that any group of order 35^3 has a normal subgroup of order 125.

    2. Relevant equations

    3. The attempt at a solution

    Is this a valid proof?

    Let G be an arbitrary group of order 353. Note that 353 = 5373. Thus, by Sylow's first theorem, there is a sylow p-subgroup of order 125, which we refer to as H. But then, by Sylow's second theorem, it follows that H is conjugate to itself in G. Hence, H is normal in G.
  2. jcsd
  3. Apr 19, 2012 #2
    I believe you are on the right track, however, I believe you have made an assumption. Sylow's second theorem says that all the Sylow 5-subgroups are conjugate to each other. This does imply normality of H iff H is the only Sylow 5-subgroup of G. I think you can use Sylow's third theorem to show that H is the only Sylow 5-subgroup.
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