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Group Theory - Sylow's Theorem

  1. Aug 30, 2007 #1
    Hey ho, just wondering if anyone knows of any decent sites which deal with Sylow's theorem.

    Twiddles :)
  2. jcsd
  3. Aug 31, 2007 #2

    Chris Hillman

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    1. Twiddles, I don't know of readable and stable and reliable websites discussing the Sylow theorems, but I can suggest some good inexpensive books, depending upon your interests and tastes. (It goes without saying that a good book is readable, stable, and reliable.) For example, do you think geometrically? If so, a good book for you might be Neumann, Stoy, and Thompson, Groups and Geometry, Oxford University Press, 1999 (see chapter 8).

    2. I suggest that a friendly moderator move this thread to the "Linear and Modern Algebra" subforum.
  4. Aug 31, 2007 #3
    How about this?
  5. Aug 31, 2007 #4


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    on my webpage, the 100 page algebra book covers them.
  6. Sep 1, 2007 #5


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    here is the standard proof of sylow, rediscovered by helmut wielandt.
    Next we use actions to produce stabilizer subgroups of prime power orders.
    Theorem(Sylow): Let #(G) = mp^r where p does not divide m.
    1) There exist subgroups of G of order p^r.
    2) All subgroups of order p^r are conjugate to one another,
    3) The number of subgroups of order p^r divides m, and is congruent to 1 modulo p.
    proof: Suppose G acts on a set S such that p does not divide #(S). S is a disjoint union of orbits, so there is an orbit O(x) whose order is not divisible by p. By the counting principle pr divides #(Stab(x)). So if we can find such an action where #(Stab(x)) is not greater than p^r, we would be done.
    Since G is an arbitrary group, the only thing G acts on is G itself, by translation, and conjugation. But G has order divisible by p. We might consider subgroups of G, but we do not know how many there are! So we consider subsets of G, with G acting by translation. If a subgroup H stabilizes a non empty set T, then for any y in T, translation is an injection H-->T taking g in H to gy in T. So H is no larger than T. Thus if we let G act on subsets of size pr, then the stabilizers will have cardinality less than or equal to p^r as desired.
    So we hope the number of such subsets is not divisible by p. Of course the set S of subsets of G of size p^r, has order given by the usual binomial coefficient . In this fraction every factor in the top of form (mp^r-k), is divisible by p^s , s less than or equal to r, if and only if k is, if and only if the factor (p^r-k) in the bottom is. Thus every factor of p occurring in the top is canceled by a factor from the bottom. Hence this binomial coefficient is not divisible by p, and thus the stabilizer of any subset in an orbit not divisible by p, gives a subgroup of G of order p^r. QED

    Lemma: If H,K are subgroups of G and H lies in N(K), then the set of products HK is a subgroup of G, and HK/K is isomorphic to H/(HmeetK).
    proof: exercise.
    To count the number of subgroups P1,...,Pn, of order p^r, (called Sylow p - subgroups, or p^r - subgroups) let P1 act by conjugation on all of them. We claim P1 fixes only P1. To prove it, if P1 fixes Pj, then P1 lies in the "normalizer" N(Pj) = {g in G such that g-1Pjg = Pj}. Then P1Pj is a subgroup of G, and (P1Pj)/Pj isomorphic to P1/(P1meetPj). Since the latter quotient group has order dividing #(P1) = p^r, it follows that #(P1Pj) is a power of p. Since P1Pj contains P1, whose order is already the largest possible power of p for a subgroup of G, hence P1 = Pj. Thus the action of P1 on the set S of Sylow p subgroups, has exactly one fixed point. By the counting principle above for p-groups, #(S) is congruent to 1, mod p.
    Now let G act on S by conjugation. The G- orbit of Pj contains the P1 orbit of Pj. Thus the G orbits are unions of P1 orbits, and all the P1 orbits except {P1}, have order divisible by p. So the G orbit containing P1 has order congruent to 1 mod p, while the others are divisible by p. But the normalizer of any Pj in G contains Pj. The order of the G orbit of Pj equals the index of that normalizer, hence divides m, so cannot be divisible by p. Thus there is only one G orbit, i.e. all Pj are conjugate. Since the order of each orbit divides m, and there is only one orbit, #(S) divides m. QED.
    Last edited: Sep 1, 2007
  7. Sep 1, 2007 #6
    Thanks guys :)

    Quick question, and this is for an assignment so just nudge me in the right direction:

    'Suppose G is a group of order 17. What can you tell me about the nature of G? Prove it.'

    I've got as far as saying that it's simple(By Lagrange's theorem). Is there anything else I should be picking up on?

  8. Sep 2, 2007 #7


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    17 is prime so there is only one group of order 17, the cyclic group.
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