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Sylow subgroups of S4

  1. Mar 26, 2012 #1
    I'm reading some notes concerning Sylow subgroups of S4:

    Since #S4 = 24 = 3 . 2^3 from Sylow's Theorems we know there are either 1 or 4 Sylow 3-subgroups and either 1 or 3 Sylow 2-subgroups.

    The question I have regards how we narrow this down:

    My notes argue that since S4 has 9 elements of order 2, they cannot all fit in a subgroup of order 8 and so we must have 3 Sylow 2-subgroups. Similarly we must have 4 Sylow 3-subgroups.

    I don't quite follow this reasoning, why must all elements of order 2 lie in a Sylow 2-subgroup? For instance, why couldn't we have some of them lying in a subgroup of order 6?

    I know you can argue that if there is only one Sylow subgroup then it must be normal and hence a union of conjugacy classes and get a contradiction that way, but I'd like to understand the logic behind the above reasoning.

  2. jcsd
  3. Mar 27, 2012 #2


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    The given argument is simply using the fact that an element of order 2 must belong to a Sylow 2-subgroup. (Because such an element generates a 2-group, which must be contained in a maximal 2-group, i.e. a Sylow 2-subgroup.) It could still belong to a subgroup of order 6, of course.
  4. Mar 29, 2012 #3
    Ah I see. Thanks for clearing that up.
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