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Group theory hints?

  1. Feb 28, 2009 #1
    Hi everyone
    This is the same question as was asked about in this topic, but I can't post in that one (presumably because it's archived?)

    1. The problem statement, all variables and given/known data

    Suppose G is a group with |G| = 4n+2. Show that there is a subgroup H < G such that |H| = 2n + 1. Use Cauchy's theorem, Cayley's theorem and the fact that any subgroup of [itex]S_n[/itex] has either all of its elements or precisely half of its elements being even.

    2. Relevant equations

    Cayley: Every group G is a subgroup of [itex]S_|_G_|[/itex].

    Cauchy: If G is a finite group & P [tex]\in[/tex] N is prime with P | |G|, then G contains an element of order P.

    3. The attempt at a solution

    Cayley [tex]\Rightarrow G \leq[/tex] [itex]S_4_n_+_2[/itex] [tex]\Rightarrow[/tex] all or half of its elements are even (*)

    Cauchy [tex]\Rightarrow[/tex] G contains an element of order 2 (since 2 | 4n+2)
    [tex]\Rightarrow[/tex] G contains an element that's a product of disjoint transpositions.

    Suppose G contains an element that's a product of an odd number of disjoint transpositions. (**)
    [tex]\Rightarrow[/tex] Half the elements in G are even, by (*), i.e. 2n+1 of them are even.
    These 2n+1 elements form a group (since they're even, their products are even, their inverses are even, and 1 is even), which is a subgroup of G with order 2n+1.

    ... So what I need to do now is show that G contains an element that's a product of an odd number of disjoint transpositions, i.e. prove that (**) is true.
    And that's where I'm stuck... I have no idea what to do.. Can I have a nudge in the right direction?
    Thanks
     
  2. jcsd
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