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Semi-Direct Product & Non-Abelian Groups

  1. Jul 28, 2011 #1
    1. The problem statement, all variables and given/known data

    Let p, q be distinct primes s.t. p [itex]\equiv[/itex] 1 (mod q). Prove that there exists a non-Abelian group of order pq and calculate the character table.


    2. Relevant equations

    Semi-Direct Product: Let H = < Y | S > and N = < X | R > be groups and let [itex]\phi[/itex] : H [itex]\rightarrow[/itex] Aut (N) be a homomorphism. Then the SDP has the presentation:

    N x H = < X, Y | R, S, [itex]y^{-1}xyw_{x,y}^{-1}[/itex] >

    With x in X, y in Y, [itex]w_{x,y}[/itex] = ([itex]\phi[/itex](y(x))) (i.e. [itex]\phi[/itex] of y and then [itex]\phi[/itex] of y of x etc.) in < Y > = N


    3. The attempt at a solution

    I've done a specific example where p=7 and q=3 and found a group of order 21, using the semi-direct product of 2 cyclic groups [itex]C_{7}[/itex] and [itex]C_{3}[/itex].

    But can't see how to prove that there exists a group more generally. So far I have.

    If N = [itex]C_{p}[/itex] = < x > and H = [itex]C_{q}[/itex] = < y >, then [itex]\phi[/itex] : [itex]C_{q}[/itex] [itex]\rightarrow[/itex] Aut ([itex]C_{p}[/itex]) = [itex]C_{p-1}[/itex] = < [itex]\alpha[/itex] >, then we have y [itex]\mapsto[/itex] [itex]\phi[/itex](y) too.

    When p=7 and q=3

    I had that [itex]\phi[/itex](y) = {1, [itex]\alpha^{2}[/itex], [itex]\alpha^{4}[/itex]}

    If [itex]\phi[/itex](y) = 1, then we'd have that the SDP of [itex]C_{7}[/itex] and [itex]C_{3}[/itex] would be ismorphic to [itex]C_{7}[/itex] x [itex]C_{3}[/itex].

    If [itex]\phi[/itex](y) = [itex]\alpha^{2}[/itex], then the SDP of [itex]C_{7}[/itex] and [itex]C_{3}[/itex] would be equal to < x, y >

    Then by the relation in the SDP, we'd have: yx[itex]y^{-1}[/itex] = [itex]\alpha^{2}[/itex](x) = x^{t}

    And we have that: t = {1, 2, 3, 4, 5, 6}, with p=7 and q=3, then: t^{3} [itex]\equiv[/itex] 1 (mod 7), and so: t = 1 and t = 2 and t = 4, thus we have groups:

    G = < X, Y | [itex]x^{7} = y^{3} = 1[/itex], [itex]yxy^{-1} = x[/itex] > = [itex]C_{7}[/itex] x [itex]C_{3}[/itex]

    G = < X, Y | [itex]x^{7} = y^{3} = 1[/itex], [itex]yxy^{-1} = x^{2}[/itex] >

    G = < X, Y | [itex]x^{7} = y^{3} = 1[/itex], [itex]yxy^{-1} = x^{4}[/itex] >

    But how would I get to this stage more generally?! Thanks for any help!
     
  2. jcsd
  3. Jul 28, 2011 #2

    micromass

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    Hi OMM! :smile:

    I think it's clear that you will want to form the group [itex]C_p\rtimes C_q[/itex]. The first thing you will want to do is check if there exists a suitable homomorphism

    [tex]\phi:C_q\rightarrow Aut(C_p)[/tex]
     
  4. Jul 28, 2011 #3
    Hi micromass, thanks for your help once again!

    So if we say, < y > = [itex]C_{q}[/itex], then we need to show there is a homomorphism [itex]\phi[/itex] that sends y to [itex]\phi[/itex](y) which is an element of Aut [itex]C_{p}[/itex] = [itex]C_{p-1}[/itex].

    And since p [itex]\equiv[/itex] 1 (mod q), then [itex]C_{p-1}[/itex] isomorphic to [itex]C_{q}[/itex]?

    And clearly there exists a homomorphism from one group to itself (endomorphism?).

    Am I on the right track?!
     
  5. Jul 28, 2011 #4

    micromass

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    You're certainly on the right track, but there is a small mistake. You say that [itex]C_q[/itex] is isomorphic to [itex]C_{p-1}[/itex], but this is not true.

    All you know is that p-1=0 (mod q), but this does not imply that q=p-1. It merely implies that q divides p-1.
     
  6. Jul 28, 2011 #5
    So for y in [itex]C_{q}[/itex], we have [itex]y^{q}[/itex] = 1

    And if q divides p-1, then [itex]y^{p-1}[/itex] = 1?

    Does the order of the element [itex]\phi[/itex](y) divide p-1? Or does it divide q?

    (Sorry, probably obvious, but having a mental moment!)

    Then to find the possible element for [itex]\phi[/itex](y) for [itex]C_{p-1}[/itex] = < [itex]\alpha[/itex] >

    [itex]\phi[/itex](y) in {1, [itex]\alpha[/itex], [itex]\alpha^{2}[/itex], .... , [itex]\alpha^{p-1}[/itex]}

    Show which powers of [itex]\alpha[/itex], when put to the power of q, give a power 1 (mod p)? (i.e. yx[itex]y^{-1}[/itex] = [itex]x^{t}[/itex], we need [itex]t^{q} \equiv[/itex] 1 (mod p))

    Am I missing something out here or is it just a case now of showing that basically t = 1 is one of these values of t, thus getting the non-abelian cyclic group [itex]C_{p}[/itex] x [itex]C_{q}[/itex] of order pq?
     
  7. Jul 28, 2011 #6

    micromass

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    Hmm, I think you're making it a bit too hard on yourself.

    We know that q divides p-1, does that mean that we can always find an element of order q in [itex]C_{p-1}[/itex]?
     
  8. Jul 28, 2011 #7
    I'd say yes (With very little conviction! Haha)

    As I said in the last post, if we have an element y in [itex]C_{q}[/itex] which is obviously of order q, then y is also in [itex]C_{p-1}[/itex]?
     
  9. Jul 28, 2011 #8

    micromass

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    Aah, yes. I seem to be a bit confused with the notation. You are identifying [itex]C_q[/itex] with elements on the unit circle of [itex]\mathbb{C}[/itex] yes??
     
  10. Jul 28, 2011 #9
    I'm not entirely sure!

    I am basically saying that C_{q} is the cyclic group of order q, thus is generated by < y >.

    In other words: C_{q} = < y | y^{q} = 1 >

    The elements of which are: {1, y, y^{2}, .... , y^{q-1}}
     
  11. Jul 28, 2011 #10

    micromass

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    Yes, that's what I thought.

    There are several methods to prove the existance of an element of order q in [itex]C_{p-1}[/itex], the easiest makes use of the fundamental theorem of abelian groups.

    However, you seem to want to do another method (which is also fine): you want to say that if y is an element of [itex]C_q[/itex], then it is an element of [itex]C_{p-1}[/itex]. Unfortunately, this does not make much sense: the elements in the cyclic group might look quite different from eachother!
    However, every cyclic group is isomorphic to a part of the unit circle by

    [tex]\Phi:C^q\rightarrow S^1:y^k\rightarrow e^{2\pi i k/q}[/tex]

    this is an group monomorphism. So now you can indeed say that an element of [itex]C_q[/itex] is also an element of [itex]C_{p-1}[/itex]!!
     
  12. Jul 28, 2011 #11
    And the order of that element divides q?

    It can't be 1, else the group would be Abelian i.e. yxy^{-1} = x ===> yx = xy

    So the order must be q, as q is prime?
     
  13. Jul 28, 2011 #12

    micromass

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    Well, you can choose that element such that the order is q!
     
  14. Jul 28, 2011 #13
    Thanks for your help, I'll give it another shot in the morning. Found a bit on Frobenius Groups which seems quite related, which I think may be my bed-time reading!

    Thanks again, you're a great help!
     
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