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Question from Allan Clark

  1. Jul 15, 2006 #1

    learningphysics

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    Stuck on this question. Question 59l (last question of section 59)

    p is a prime such that p^k divides the order of a group G. Prove that the number of subgroups of order p^k is congruent to 1 mod p. Clark warns that the solution is lengthy, and it is a theorem by Frobenius.

    I'd appreciate any help or hints! Thanks.
     
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  3. Jul 15, 2006 #2

    mathwonk

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    reminds me of a fact about group actions, namely when a group of order p^n acts on a finite set, the number of fixed points is congruent to the number of elements of the set, mod p. so look for some action on the set of sylow subgroups by one of them that has 1 fixed point?

    but make sure that in your notation, p^(k+1) does not divide the order of the group.
     
  4. Jul 15, 2006 #3

    learningphysics

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    EDIT
    Thanks for the reply mathwonk! I will try to proceed by this method.
     
    Last edited: Jul 15, 2006
  5. Jul 15, 2006 #4

    mathwonk

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    you will need to use the second sylow theorem, twice, once to get the action, and once to show the number of fixed points is 1.
     
  6. Jul 15, 2006 #5

    mathwonk

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    oh and the solution is not at all lengthy with this outline.
     
  7. Jul 15, 2006 #6

    learningphysics

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    mathwonk, I was a little worried when you said p^(k+1) does not divide the order of the group... Is the outline for proving that the number of psubgroups (p-sylow or not) of a particular order is 1 mod p?

    I was thinking that the outline you mentioned might be just for p-sylow groups.
     
  8. Jul 15, 2006 #7

    learningphysics

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    I have a feeling what I do here is all wrong... suppose we have a group G, where p^k divides the order of the group... we're trying to find the number of subgroups of order p^k. Let H be one of the subgroups of order p^k. Let S be the set of all subgroups of G of order p^k.

    So if we define the group action as:

    h * s = hs (h is an element of H, and s is an element of S), then we have only one fixed point, namely H...

    h * H = H because h belongs to H...

    but for any other set s in S other than H... we can find at least one h1 in H such that h1 does not belong to s (since both sets have the same order, and they are not equal)... so h1 * s does not equal s, where h1 is in H, and h1 is not in s... so s isn't a fixed point.

    so 1 is congruent to |S| mod p, and therefore |S| is congruent to 1 mod p...

    Is this alright?
     
  9. Jul 15, 2006 #8

    StatusX

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    That isn't an action on S. If by hs you mean left multiplying a subgroup s by an element h of H, then hs is not a subgroup, and so doesn't belong to S. For one thing, it doesn't have the identity. If you mean conjugation by h, then there may be other fixed points besides H.
     
  10. Jul 15, 2006 #9

    learningphysics

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    Ah... thanks StatusX.
     
  11. Jul 15, 2006 #10

    mathwonk

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    i was giving the outline for a proof for thw sylow subgroups, but as you said, the problem is for all subgroups of order a fixed power oof p.

    but it cant hurt to understand the maximal power acse first.

    let a sylow subgroup H act on the others by conjugation, instead of translation. then H fixes itself.

    claim H does not fix any other sylow subgroup: proof: if K is another one then consider N = the normalizer of K. By definition K is normal in N, so by the second sylow thm there cn be no other such subgroup in there, so H does not normalize K.

    that says the action by H on the set of sylow subgroups has exactly one fixed point, namely H itself.

    so the general principle says the number of sylow subgroups, is congruent to 1 mod p.

    this proof seems nicer to me than clarks proof, and may generalize better to a proof of frobenius thm.
     
  12. Jul 15, 2006 #11

    learningphysics

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    Thanks a lot mathwonk! Yes, that's a much better proof. I'll try to show that H only fixes itself even when we're not using sylow groups...
     
  13. Jul 15, 2006 #12

    mathwonk

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    and if that isn't true try to extend the argument somehow.
     
  14. Jul 17, 2006 #13

    mathwonk

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    i propose the following outline;
    1) show it is true for any group G whose order is a power of p.
    2) show that for each subgroup H of G, where H has order a power of p, that the number of sylow subgroups of G that H is a subset of, also is congruent to 1 mod p
    3) show that does it.
     
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