1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A Sylow Subgroup Question

  1. Nov 2, 2008 #1
    The problem statement, all variables and given/known data
    If G is a group of order 231, prove that the 11-Sylow subgroup is in the center of G.

    The attempt at a solution
    The number of 11-Sylow subgroups is 1 + 11k and this number must be either 1, 3, 7, 21, 33, 77 or 231. Upon inspection, the only possibility is 1.

    Let H be the 11-Sylow subgroup of G. Note that 3 * 7 * 11 = 231, so o(H) = 11 and H is cyclic. Furthermore, this is the only subgroup of order 11, so it must be normal. With these facts in mind, let a be a nonidentity element of H. For any g in G, we must have g-1ag = aj for some j in {1, 2, ..., 11}. Thus ag = gaj.

    I don't know what to do next. Any tips?
     
  2. jcsd
  3. Nov 3, 2008 #2

    morphism

    User Avatar
    Science Advisor
    Homework Helper

    What can you say about the order of the automorphism g -> g-1ag? (Hint: it's enough to consider those g's coming out of the 3- and 7-Sylows.)
     
  4. Nov 3, 2008 #3
    If the order of g is 3 or 7, then the order of the automorphism must be 21. But now what?
     
  5. Nov 3, 2008 #4

    morphism

    User Avatar
    Science Advisor
    Homework Helper

    Why do you say that?

    By the way, there was a typo and an unfortunate choice of letters in my previous post. But I hope the idea got through though - we want to let G act on H by conjugation. That is, consider the automorphism f_g of H defined by f_g(x) = g-1xg. What is its order? (Hint: the map g->f_g is a homomorphism.)
     
  6. Nov 4, 2008 #5
    I said 21 because because it is the least common multiple of 3 and 7. Thus, applying f_g 21 times to x will surely result in just x.
     
  7. Nov 4, 2008 #6

    morphism

    User Avatar
    Science Advisor
    Homework Helper

    Sure, but does that mean the order of f_g is 21?

    Here's another hint. If the order of g is 3, then the map g->f_g is an embedding from Z_3 into Aut(H)=~Z_10.
     
  8. Nov 4, 2008 #7
    No. Let n be the order of fg. Then g-nagn = a or agn = gna. If g in H, then n must be 1. If g is in a 3-Sylow subgroup, then n must be 3 unless g commutes with elements from H. Is this possible?

    Wait. Where did you get that Aut(H) =~ Z10? I thought Aut(H) =~ U11.
     
  9. Nov 5, 2008 #8

    morphism

    User Avatar
    Science Advisor
    Homework Helper

    If U_11 is the group of invertibles of Z_11, then it's isomorphic to Z_10 (it's cyclic and has order phi(11)=10).
     
  10. Nov 5, 2008 #9
    Oh yeah. Still, I don't see where you're going with this.
     
  11. Nov 5, 2008 #10

    morphism

    User Avatar
    Science Advisor
    Homework Helper

    Well, a subgroup is central iff conjugating it is a trivial action. Since H itself is abelian, we only need to worry about conjugation with elements outside of H. By Lagrange, the non-identity elements in G\H will lie in either the 3-Sylow or 7-Sylow.

    Let's consider the 3-Sylow, call it P. The map g -> f_g is a homomorphism from P into Aut(H)=Z_11. An order argument will yield that f_g=1. Do the same for the 7-Sylow.
     
  12. Nov 6, 2008 #11
    But the elements of G/H are cosets. I don't get.

    You meant to write Z_10. I understand know. I did not suspect the order of f_g to be 1. I'll have to think about that.
     
  13. Nov 6, 2008 #12
    I just realized you meant G - H when you wrote G\H. Sorry about that. Still, I don't see how Lagrange implies that though.
     
  14. Nov 6, 2008 #13
    Since Aut(H) is isomorphic to Z10, it has 10 elements, whence the order of fg must be either 1, 2, 5 or 10. And since we know that (fg)3 is the identity map (assuming g is an element of a 3-Sylow subgroup), we can rule out 2, 5, and 10, so its order must be 1.
     
  15. Nov 6, 2008 #14

    morphism

    User Avatar
    Science Advisor
    Homework Helper

    Yup.

    You can disregard this bit:
    It's obviously nonsense. I don't know why I wrote it.

    Anyway, to finish the argument, notice that f_g=1 for every g in P means that P is a subset of the centralizer of H. So...
     
  16. Nov 7, 2008 #15
    I understand the rest. Thank you. However, we still have to show that every non-identity element in G - H has order 3 or order 7.
     
  17. Nov 7, 2008 #16

    morphism

    User Avatar
    Science Advisor
    Homework Helper

    No we don't. It's actually not true (think of an abelian counterexample). Initially I 'broke up' the group into elements of order 3, 7 and 11 to guide my intuition, but then I took it too far.

    The point is, we can embed subgroups of order 3, 7 and 11 into the centralizer of H. So the centralizer must be the entire group. That's all there is to it.
     
  18. Nov 7, 2008 #17
    I'm not sure I understand what you mean by "embed" or how that proves that C(H) = G. We have shown that if g has order 3 or 7, then it belongs to C(H). Does the same argument we used to show that the order of fg is 1 work for those g that have order 21, 33, or 77? This seems to be the case.
     
  19. Nov 9, 2008 #18
    I've been thinking, if f_g is an automorphism of H, then g must be in H and we can't really use it if g is not in H, which destroys the whole argument we've developed.
     
  20. Nov 9, 2008 #19
    Oh wait. I forgot that H is normal so g-1xg is in H for all x in H and g in G.
     
  21. Nov 9, 2008 #20
    OK. I've written down a solution which I'm content with. Thanks for the help morphism.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: A Sylow Subgroup Question
  1. Sylow Subgroups (Replies: 1)

  2. P-Sylow subgroup (Replies: 0)

Loading...