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Conjugate Subgroups of a Finite Group

  1. Jul 30, 2014 #1
    1. The problem statement, all variables and given/known data
    Two subgroups of G, H and K are conjugate if an element a in G exists such that aHa^-1= {aha^-1|elements h in H}= K

    Prove that if G is finite, then the number of subgroups conjugate to H equals |G|/|A|.


    2. Relevant equations
    A={elements a in G|aHa^-1=H}

    3. The attempt at a solution

    Stabilizer of H also known as NG(H) is {a: aHa^-1 = H}.
    By Lagrange's Theorem, number of left cosets in H = |G|/|H|.
    NG (H) is a normalizer and NG (H) = {elements g in G| gHg^-1 = H}.
    A is normalizer and A= NG (H) because A={elements a in G|aHa^-1=H}.
    Therefore, no. of subgroups conjugate to H, is |G|/|A|.

    This argument is terribly weak and I have no idea how to formalize the proof. I want to just use the concept of subgroup/group properties not normalizers/stabilizers/centralizers because I clearly do not understand the latter well enough to do so.

    I appreciate any help. Thank you.
     
  2. jcsd
  3. Jul 30, 2014 #2

    jbunniii

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    Try defining a mapping from ##G/A## (the set of cosets of ##A##) to the set of conjugates of ##H##. The natural thing to try is to map the coset ##xA## to the conjugate ##xHx^{-1}##. Can you show that this is a well-defined mapping, i.e., if ##xA = yA## then ##xHx^{-1} = yHy^{-1}##? Is the mapping a bijection?
     
  4. Jul 30, 2014 #3
    ψ: G/A -> H, where G/A = {elements x, y in A| xA=Ax and yA=Ay} and H = {elements x, y in H|xHx^-1 and yHy^-1 = H}
    I do not think the mapping is defined correctly.
     
  5. Jul 30, 2014 #4

    jbunniii

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    No, you've created new definitions for G/A and H, but they are already defined!

    G/A is the set of (left) cosets of A. In other words, ##G/A = \{xA | x \in G\}##.

    ##H## is a subgroup of ##G## as given in the problem statement. Let's define ##X = \{xHx^{-1} | x \in G\}##, the set of conjugates of ##H##.

    My suggestion is to define ##\psi : G/A \rightarrow X## by ##\psi(xA) = xHx^{-1}##.

    If you can show that ##\psi## is a well-defined bijection, then we can conclude that the size of ##G/A## must be equal to the size of ##X##, which is the result you need.

    So you need to show:
    1. ##\psi## is well defined, i.e. if ##xA = yA## then ##xHx^{-1} = yHy^{-1}##
    2. ##\psi## is a bijection
     
  6. Jul 30, 2014 #5
    1. xA= yA if x^-1y is in H
    A = x^-1y

    Similarly, Ax^-1y= Ay^-1 if x^-1y is in H. Again, A= x^-1y.
    Then, the mapping is defined.

    2. (xA)^-1 = (yA)^-1
    (xA) (xA)^-1 = (xA) ((yA)^-1) e = (xA) ((yA)^-1)
    e(yA) = (xA)((yA)^-1)(yA)= yA = xA
    So, (xA)^-1 = (yA)^-1 <---> yA = xA
    The homomorphism is bijective.

    Therefore, if xA has order n in G/A, then xHx^-1 in X also has order n.
    Since G/A and X have the same order, and X is the set of conjugates on H, then |X| = |G|/|A|.

    Is this accurate?
     
    Last edited: Jul 30, 2014
  7. Jul 30, 2014 #6

    jbunniii

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    You mean ##x^{-1}y \in A##, not ##H##. Also, this is an "if and only if", which will be useful later.
    No, this makes no sense. The left hand side is a set of elements (indeed, a subgroup of ##G##), whereas the right hand side is a single element.

    You mean "is in ##A##".
    Again, this makes no sense.

    The goal is to show that if ##xA = yA##, then ##xHx^{-1} = yHy^{-1}##. So assume ##xA = yA##. Thus ##x## and ##y## are in the same coset of ##A##. Therefore ##y^{-1}x \in A##. By definition of ##A##, this means that ##(y^{-1}x)H(y^{-1}x)^{-1} = H##. Can you finish the proof from here?

    I have no idea what you're doing here. You need to show that ##\psi## is a bijection. (By the way, it won't be a homomorphism.) To show that ##\psi## is a bijection, you need to show that it is an injection and a surjection. Let's show that it is an injection. To do this, we need to show that if ##xHx^{-1}= yHy^{-1}## then ##xA = yA##. So suppose that ##xHx^{-1} = yHy^{-1}##. Multiplying on the left by ##y^{-1}## and on the right by ##y##, we get ##y^{-1}xHx^{-1}y = H##. Can you finish the proof?
     
    Last edited: Jul 30, 2014
  8. Jul 30, 2014 #7
    If xy^-1Hyx^-1 = H, and y^-1Hy = H then x(y^-1Hy)x^-1 = xHx^-1. So, when xHx^-1 = H = H, the mapping of H (onto itself) is well defined.


    If y^-1xHx^-1y = H, and H = xHx^-1 , then y^-1Hy = H and H = H?
     
  9. Jul 30, 2014 #8

    jbunniii

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    No, you cannot assume that ##y^{-1}Hy = H##. It won't be true in general. If you go back to what I wrote above, I showed that if ##xA = yA##, then ##(y^{-1}x)H(y^{-1}x)^{-1} = H##. The goal is to show that this implies that ##xHx^{-1} = yHy^{-1}##. So you need to show how to go from
    $$(y^{-1}x)H(y^{-1}x)^{-1} = H$$
    to
    $$xHx^{-1} = yHy^{-1}$$
    Same problem here. You can't assume that ##H = xHx^{-1}##. It is not true in general. You need to show that if ##y^{-1}xHx^{-1}y = H##, then ##xA = yA##. I'll do another step for you. Given that ##y^{-1}xHx^{-1}y = H##, we can rewrite this slightly as
    $$y^{-1}xH(y^{-1}x)^{-1} = H$$
    Therefore, ##y^{-1}x \in A##. (Why?)
     
  10. Jul 30, 2014 #9
    Then (y^-1x)H(x^-1y) = H and xHx^-1 = yHy^-1 by multiplication.

    y^-1x is contained in A because if x and y are in the same coset of A, and xy^-1 is contained in A, then y^-1x should also be contained in A. (Is x ~ y an equivalence relation in A?)
     
  11. Jul 30, 2014 #10

    jbunniii

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    Looks good.
    No, this is arguing circularly. You need to explain why
    $$y^{-1}xH(y^{-1}x)^{-1} = H$$
    implies
    $$y^{-1}x \in A$$
    What is the definition of ##A##?
     
  12. Jul 30, 2014 #11
    A={elements a in G|aHa^-1=H}

    I do not understand if we cannot use the assumption that xHx^-1 = H then how can we come to the conclusion that y^-1x is in A?
     
  13. Jul 30, 2014 #12

    jbunniii

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    Look again at ##y^{-1}xH(y^{-1}x)^{-1} = H##. If I define ##a = y^{-1}x##, then this equation becomes
    $$aHa^{-1} = H$$
    Therefore...? What can you say about ##a##?
     
  14. Jul 30, 2014 #13
    Then element a is contained in A (and H).
     
  15. Jul 30, 2014 #14

    jbunniii

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    Yes. So what can you conclude?
    Not necessarily!
     
  16. Jul 30, 2014 #15
    Oh I realize my mistake. H is a subgroup of A?

    If a is an element in A, then y^-1x is also contained in A. Then xA = yA?
     
  17. Jul 30, 2014 #16

    jbunniii

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    Yes, that's correct. The following relation holds: ##H \unlhd A \leq G## (in words, ##H## is a normal subgroup of ##A##, and ##A## is a subgroup of ##G##). Incidentally, ##A## is called the normalizer of ##H##, because it is the largest subgroup of ##G## in which ##H## is normal.
    Right, ##a = y^{-1}x##, so ##y^{-1}x \in A##. Therefore, ##x \in yA##, in other words, ##x## and ##y## are in the same coset of ##A##, in other words, ##xA = yA##.
     
  18. Jul 30, 2014 #17
    And now because ψ is well defined and isomorphic, we determine that |G|/|A| = |X|.
     
  19. Jul 30, 2014 #18

    jbunniii

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    Almost correct. In fact, ##\psi## is a bijection but it is not necessarily a homomorphism. (Indeed, ##G/A## is not even a group unless ##A## happens to be normal in ##G##.) But if you change one word, it's correct:

    And now because ψ is well defined and bijective, we determine that |G|/|A| = |X|.

    (Here we have used the fact that ##|G/A| = |G|/|A|## by Lagrange's theorem.)

    By the way, you have shown that ##\psi## is an injection, but you haven't yet shown that is a surjection. Fortunately, that part is easy. Can you see why it's true?
     
  20. Jul 31, 2014 #19
    For every xA in G/A, there must be an xHx^-1 in X. This is explained by the mapping of Psi(xA) onto xHx^-1, therefore the mapping is surjective as well?
     
  21. Jul 31, 2014 #20

    jbunniii

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    The goal is to show that ##\psi : G/A \rightarrow X## is a surjection. Therefore your proof should start with "for every ##xHx^{-1} \in X##, the element ____ of ##G/A## satisfies ##\psi(##___##) = xHx^{-1}##." (Fill in the blank.)
     
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