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Application of Sylow's theorems

  1. Oct 14, 2009 #1
    I saw in an application of Sylow's theorems, it said we have something like a group of order 28 = 2^2 x 7, so we have either 1 or 7 sylow 2-subgroup. Assuming we have 7 sylow 2-subgroups, then we have 21 non-identity elements and the identity, and we have 1 sylow 7-subgroup, blah blah blah....

    the point I wanted to know that was never really explained clearly was that why are the 7 sylow 2-subgroups intersection trivial? I realize that they are conjugates of each other, but this counting of elements seems kind of moot unless we know that these subgroups don't have any common elements.
  2. jcsd
  3. Oct 15, 2009 #2


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    Re: Sylow

    Sylow subgroups do NOT in general have trivial intersection. What you can say, if you have 7 Sylow 2-subgroups, is that their union is at MOST 21 non-identity elements plus the identity.
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