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The product of Sylow subgroups

  1. Jan 28, 2012 #1
    If |G|=(p_1)^s1 (p_2)^s2 ...(p_t)^st where p_i are primes and si are positive integers then
    G=P_1 P_2 ... P_t where P_i are the Sylow subgroups of G.
    Is this true?
    I think it is right because G and P_1 P_2 ... P_t have the same order. Please, help me.
     
  2. jcsd
  3. Jan 28, 2012 #2

    morphism

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    It's not true. The problem is that the "product" P_1 P_2 ... P_t of subgroups is not well-defined in general. This is because: given arbitrary p_i, q_i \in P_i, there is no reason for their product (p_1 ... p_t)(q_1 ... q_t) to be expressible in the form r_1 ... r_t with r_i in P_i.

    Let's suppose we are in the nicest case possible where the internal direct product P_1 ... P_t of all the Sylow subgroups makes sense. Then this forces the group G to be nilpotent, and the converse is true: every finite nilpotent group is the direct product of its Sylow subgroups.
     
  4. Jan 28, 2012 #3

    Deveno

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    an example might be illustrative, here.

    suppose G = S3, the symmetric group on 3 letters (this is a nice, small group). we can take P2 to be <(1 2)>, and P3 to be <(1 2 3)>.

    now, it is true that the product SET:

    P2P3 = {xy in S3: x in P2, y in P3} is indeed equal to S3.

    the trouble lies with how to define a multiplication on P2P3, solely in terms of the multiplications IN P2, P3.

    to see that the "expected" product (xy)*(x'y') = (xx'yy') doesn't work, suppose:

    x = x' = (1 2)
    y = (1 2 3), y' = (1 3 2).

    then xx'yy' = (1 2)(1 2)(1 2 3)(1 3 2) = ee = e.

    but, in fact, (1 2)(1 2 3)(1 2)(1 3 2) = (2 3)(1 3) = (1 2 3), not e.

    this happens because the 2-sylow subgroup(s) of S3 are not normal in S3, so S3 is not a direct product of its sylow subgroups.

    now, for some groups, there is a way around this (called a semi-direct product), but the multiplication's definition is rather involved.
     
  5. Jan 29, 2012 #4
    If I take the product set of the Sylow subgroups of G and the operation of G as in S_3 example above, then this is not a subgroup of G. In my question, I did not mean the internal direct product, just the product set with the operation of G.
    Why do I need to define a new multiplication?
     
    Last edited: Jan 29, 2012
  6. Jan 29, 2012 #5

    morphism

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    In that case yes, G is equal to P_1 ... P_t.
     
  7. Jan 29, 2012 #6
    Thank you very much for your help.
     
  8. Jan 29, 2012 #7

    Deveno

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    you don't....if you already know a way to compute products in G, that works.

    but...suppose the order of G is, say, 48. the full multiplication table for G (if G is very non-abelian) could require 2,304 computations.

    on the other hand, if you can reduce G to a direct product of a group of order 16, and a group of order 3, that reduces the number of computations to at most 262 (256 for the worst case of a non-abelian group of order 16, and 6 more for the table of the cyclic group of order 3).

    this savings is still realized (in part) even with a semi-direct product (although there is extra calculation involved in determining the automorphisms of the normal sylow subgroup).

    this is one reason why we "decompose" groups into smaller groups, the smaller groups are easier to deal with (and often have "nicer" properties than the entire group...they may be cyclic, or perhaps abelian, or perhaps solvable). it's sort of like factoring integers into primes. if we know enough about the prime factors, that can often tell us a great deal about the big "composite" number, without a lot of tedious arithmetic.

    morhpism's post is to the point: nilpotent groups are well worth studying, because for those groups, the sylow subgroups basically tell us "everything we need to know". the interactions of the elements of the various sylow subgroups "don't interfere with each other". this makes them "nicer than some groups", we can study them "one prime factor (of the order of G) at a time", in other words, we've reduced the study of nipotent groups, to the study of p-groups (which have a whole host of properties "typical" groups do not).

    your observation is correct, but it is not terribly useful. how many conjugates a sylow subgroup has, turns out to be fairly important for understanding a given group's behavior. normal is best, it allows us to factor out that subgroup, and delve more into the mystery of what remains.

    one of the consequences of this is, for abelian groups, we can describe them ENTIRELY as direct products of cyclic groups (cylic p-groups, in fact). in effect, this makes abelian groups really just "linear combinations of integers/integers mod pk" (perhaps you can see how some of the basic structure for linear algebra starts to emerge out of this). this means abelian groups are "user-friendly", if you can add and multiply integers, you can do the math required for abelian groups (with a few extra wrinkles). this lets us describe quite a lot of what might seem like abstract structures by counting things. and counting is easy.
     
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