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Groups of order 60 and elements of order 5

  1. Aug 10, 2007 #1
    1. The problem statement, all variables and given/known data

    Let G be a group with order [tex]\left| G \right| = 60[/tex]. Assume that G is simple.

    Now let H be the set of all elements that can be written as a product of elements of order 5 in G. Show that H is a normal subgroup of G. Then conclude that H = G

    2. Relevant equations

    3. The attempt at a solution

    I started by proving that H acutally is a subgroup.

    I've then shown that there are 6 Sylow-5 subgroups in G and that they are cyclic. I know that all the elements of order 5 are the generators of the Sylow-5 subgroups. But how I can use that to show that H is normal escapes me.

    All help/hints appreciated :).
  2. jcsd
  3. Aug 10, 2007 #2

    matt grime

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    Sylow, won't help you, I don't think - the elements of order 5 do not generate Sylow-5 subgroups. The product of two elements of order 5 is not necessarily an element of order 5 (or any power of 5).

    H is trivially a subgroup - there is nothing to prove there. What about normality? This is straight forward - conjugation preserves order, and notice that

    gxyg^{-1} = gxg^{1-}gyg^{-1}
  4. Aug 10, 2007 #3


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    Count the total number of elements in all the 5-sylow groups. Use that number to show that the subgroup must be normal
  5. Aug 10, 2007 #4
    Thanks for the quick help. I understand the solution now o:).
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