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Sylow's Counting Argument

  1. Feb 22, 2008 #1
    Hi last one here. Any hints on this is appreciated too :)

    Let G be a group of order 44. Show using Sylow's counting that G has a normal subgroup of order 11. Use the results to classify all groups of order 44.
  2. jcsd
  3. Feb 22, 2008 #2


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    How many subgroups of order 11 are there? (There are a bunch of other threads about Sylow counting arguments just like this one.)
  4. Feb 23, 2008 #3
    ok here is what i have managed to get so far

    By Sylow's 1st theorem, G has a subgroup of order 11. let [itex]n_p[/itex] be the number of sylow p-subgroups. then [itex]n_{11} = 1(mod 11)[/itex] and [itex]n_{11}[/itex] divides 2^2 (=4) so [itex]n_{11}=1[/itex]. Therefore, it must be a normal subgroup (since it has no distinct conjugates).

    How can we use this to classify all groups of order 44 though?
  5. Feb 23, 2008 #4


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    Well, what are your thoughts? The obvious thing is to think about the Sylow 2s. There's either 1 or 11 of them (why?). If there's just 1, say H, then G is isomorphic to the direct product of H and the Sylow 11 (why?). Now what?

    Scenario 2: there's 11 Sylow 2s. Take one of them and play around with it and with the Sylow 11. Remember that the Sylow 11 is cyclic and normal in G -- this will probably be useful in getting a presentation for G.
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