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Abelian subgroup of GL(n, C)

  1. Jul 9, 2007 #1
    How do finite abelian subgroups of GL(n, C) with n > 1 look like ?
    I would say the elements of those subgroups are only the diagonal matrixes but Im not sure (for my homework I do not have to prove it but I want to use this result if it is true).

    GL(n, C) are all the invertible matrixes over the complex numbers of size n by n.
     
    Last edited: Jul 10, 2007
  2. jcsd
  3. Jul 9, 2007 #2

    matt grime

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    Your conjecture is clearly false. Pick any non-diagonal matrix: it generates an abelian subgroup. (My first response was answering the wrong question - I hope you didn't catch it in the 5 seconds it was posted.)

    and of course, given any subgroup, any conjugate set of matrices is an isomorphic subgroup, and diagonal matrices are not closed under conjugation (only scalar ones, where all diagonal entries are the same are).
     
  4. Jul 9, 2007 #3
    wait a minute, one of my algebra books mention this as a theorem I think. I ll have a look at it.

    edit: well, they only say something about the center.
     
  5. Jul 9, 2007 #4
    hmm, thanks for the info Matt. I thought I tackled the original question but I have to look at it again Im afraid.
     
  6. Jul 10, 2007 #5
    Big appologies but the abelian subgroup has to be finite.
    In that case I do really think it is true
     
  7. Jul 10, 2007 #6

    matt grime

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    Again that is clearly false. Find some simple examples, like C_2, the group with 2 elements. It's even false over the integers, never mind C.

    The center of GL(C,n) is the set of scalar matrices, that is the result you're thinking of.
     
  8. Jul 10, 2007 #7
    :frown:

    an easy counterexample is indeed the group generated by the square matrix

    Code (Text):
    1  0
    1 -1
     
  9. Jul 10, 2007 #8

    NateTG

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    I wonder if thinking in terms of eigenvectors and eigenvalues will lead to a complete solution.
     
  10. Jul 10, 2007 #9

    matt grime

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    Perhaps...

    look, any finite abelian subgroup of GL(n,C) is a representation of that group - this is completely reducible, hence all the matrices are simultaneously diagonalizable.
     
  11. Jul 12, 2007 #10
    I got a result in my book which says Im partly right.

    Given an irreducible representation X : G -> GL(n, C). Suppose that a square matrix M commutes with X(g) for every g in G. Then M = cI with c a scalar.

    I can use this very well in my homework.

    thanks for your help anyway Matt. Appreciated
     
  12. Jul 12, 2007 #11

    matt grime

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    in what sense?


    This is Schur's lemma, and has nothing to do with your original question, as far as I can tell, which asked what the abelian subgroups of GL(n,C) are. As I told you, they are subgroups *conjugate* to diagonal matrices.
     
  13. Jul 12, 2007 #12
    yes but now in my exercise I have that the elements of the image are of the form cI for a scalar c in C. Exactly what I want cause now I got my contradiction.
     
  14. Jul 12, 2007 #13

    matt grime

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    What? I am confused. The only map you wrote down was an irrep X:G-->GL(n,C). It is certainly not the case that the image of X must be in the scalar matrices. Note that if you restrict G to being abelian then the only irreducible representations are 1-dimensional (i.e. n=1 necessarily).

    Please, state clearly what you want to show, and what you think you have shown.
     
    Last edited: Jul 12, 2007
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