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Math Help - Sylow's Theorem

  1. #1
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    Sylow's Theorem

    Let |G|=p^{\alpha}m, \ p\nmid m. Then Syl_p(G)\neq\O that is n_p\geq 1 where n_p number of Sylow p subgroups.

    Let G be a minimal counterexample.

    p\mid |Z(G)| \ \ \text{or} \ \ p\nmid |Z(G)|

    Why are me looking at p dividing or not dividing the center of G?
    Last edited by dwsmith; November 12th 2011 at 08:18 PM.
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  2. #2
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    Re: Sylow's Theorem

    Quote Originally Posted by dwsmith View Post
    Let |G|=p^{\alpha}m, \ p\nmid m. Then Sly_p(G)\neq\O that is n_p\geq 1 where n_p=|Sly_p(G)|.

    Let G be a minimal counterexample.

    p\mid |Z(G)| \ \ \text{or} \ \ p\nmid |Z(G)|

    Why are we looking at p dividing or not dividing the center of G?
    what is your working definition of Syl_p(G)? also, is n_p supposed to number of subgroups Syl_p(G), or the cardinality of such a subgroup?

    without having a bit more context, i can only hazard a guess, but my thought is, you are looking at an inductive proof, and will eventually consider G/Z(G).
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  3. #3
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    Re: Sylow's Theorem

    I will just type up the whole proof but my question is still in regards to the center.

    proof:
    Let G be a minimal counterexample. Either p||Z(G)| \ \text{or} \ p\nmid |Z(G)|.

    Case 1: p||Z(G)|
    The center is abelian. By Cauchy's Theorem (Let G be a finite abelian group and p||Z(G)|, then G has an element of order p), there is an element g\in Z(G) of order p.
    Set <g>=H. H\trianglelefteq G since g\in Z(G).
    |G|=|G/H||H|=|G/H|p since the order of H is p.
    |G|=p^{\alpha}m=|G/H|p
    So |G/H|<|G|.
    Since G is a minimal counterexample, G\H has a subgroup of order p^{\alpha-1} (WHY), say K<G\H.
    By the 4th Isomorphism Theorem, there exist N\leq G such that H\leq N and N/H=K.
    Then |N/H||H|=|N| which is p^{\alpha-1}p=p^{\alpha}.
    Therefore, N\in Syl_p(G)

    Case 2: p\nmid |Z(G)|
    By the Class Equation, p^{\alpha}m=|G|=|Z(G)|+\sum_{i=1}^r|G:C_G(g_i)|.
    So |Z(G)|=|G|-\sum_{i=1}^r|G:C_G(g_i)|.
    p must not divide |G:C_G(g_i)| for some i (WHY). By Lagrange's Theorem, |G:C_G(g_i)||C_G(g_i)|=|G|. Since p divides G and by Euclid's Lemma, p||C_G(g_i)|; moreover, p^{\alpha}||C_G(g_i)|.
    So |C_G(g_i)|=p^{\alpha}m' where m'<m otherwise G=C_G(g_i) which contradicts the Class Equation.
    By induction, there exist P\leq C_G(g_i) such that |P|=p^{\alpha} so P\in Syl_p(G)
    Last edited by dwsmith; November 12th 2011 at 08:26 PM.
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  4. #4
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    Re: Sylow's Theorem

    ok, now we have some context.

    the reason why we look at |Z(G)|, is in order to use the class equation.

    specifically, if p does not divide |Z(G)|, then the sum of the centralizer indexes must not be divisible by p, either. because if the sum was, then

    since p divides |G|, and the sum of centralizer indexes, it would divide the difference, which is |Z(G)|, a contradiction.

    in turn, if the sum of centralizer indexes is not divisible by p, at least one centralizer index is not divisible by p.

    but since |G| = [G:C(g)]|C(g)| for any centralizer C(g), all the powers of p in the factorization of |G| are in |C(g)|.

    since C(g) is strictly less than G (since otherwise g is in Z(G)), it has a p-sylow subgroup, which in turn is a subgroup of G.

    however, if p divides Z(G), we can't use the class equation, because it might be that p divides the index of all our centralizers.

    so then we use cauchy's theorem for abelian groups to produce a normal subgroup of G, <g>, of order p, and consider G/<g>.

    since |G/<g>| < |G|, we can assume G/<g> has a p-sylow subgroup (since G is, by assumption, the minimal counter-example).

    this subgroup has order p^{\alpha-1}. we then use the 1-1 correspondence of subgroups of G/<g> and subgroups of G

    containing <g> (guaranteed by the 4th isomorphism theorem, which is a consequence of the 1st isomorphism theorem),

    to produce a subgroup of G containing <g> which has order p^{\alpha-1} (this subgroup is N in the proof).

    N/<g> is isomorphic to the p-sylow subgroup of G/<g>, and |N| = |N/<g>||<g>| = (p^{\alpha-1})(p) = p^\alpha,

    which is the desired p-sylow subgroup of G.

    in either case, we get a contradiction: by assuming G is a minimal counter-example, we prove G is NOT a counter-example.

    therefore, there is no minimal counter-example, so every finite group G has a p-sylow subgroup.
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