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Centralizer question

  1. Jun 1, 2010 #1
    1. The problem statement, all variables and given/known data
    Assume that G is a group of order 63 that has two Sylow subgroups whose intersecion is non-trivial. Show that G has an element of order 21.



    2. Relevant equations



    3. The attempt at a solution
    So by Sylow's theroem, I know that 63=3^2*7, that the sylow 7 subgroup is a normal subgroup of G with 6 order 7 elements. My sylow 3 subgroup is of order 9 and there can be one or 7 sylow three subgroups. Since I know that s7 (sylow 7 subgroup) is normal.

    So since we have two sylow subgroups whose intersection is non-trival, ths means that we have 7 sylow 3 subgroups. This means we have 14 elements of order 3 in are group G. Since we have a unique sylow 7 subgroup, we have 6 elements of order 7 in G.
    Let H = sylow 7 subgroup.
    So for every a in G, aha^-1 is a subgroup of order 7, so we must have aHa^-1=H (since H is normal). So N(H) = G (N(H)= normalizer of H). SInce H has prime order, it is cyclic and abelian. So C(H) contains H (C(H)= centralizer of H in G). So 7 divides |C(H)| and |C(H)| divides 63. So C(H)= H or C(H)=G, or |C(H)|= 21.
    Is this enough to get that xy=yx wth |x,y|=21?
     
  2. jcsd
  3. Jun 2, 2010 #2
    I think you are on the right track. You need to show that there is an element x of order 3 that commutes with an element y of order 7, then z=xy would have order 21. If you can show C(H) has order 21 (or 63) then you are done (by Cauchy's theorem).

    So all you need to do is show that C(H) is not equal to H. As you say, N(H)=G. There is a theorem that says N(H)/C(H) is isomorphic to a subgroup of the automorphism group of H. H is cyclic of order 7, so its automorphism group has order 6. So N(H)/C(H) cannot have order 7.
     
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