1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: 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
    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
    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
    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
    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
    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
    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!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook