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Commutant of GL(n = 2, Z_p)

  1. Oct 13, 2007 #1
    What is it? I have shown that the groups GL(n = 2, Z_2) and D_3, the dihedral group with order 6, are isomorphic, so now I find myself wondering if the groups GL(n = 2, Z_p) and the corresponding dihedral group with the same order, are isomorphic. That way I could find the commutant by looking at the commutant for the dihedral group instead which is much easier to get.
     
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  3. Oct 13, 2007 #2
    I've noticed now that the order of GL for p = 3 is an odd number for which can't possibly be the order of any dihedral group. So what I said might work only for even orders of GL. Anyways I still don't know how to get the commutant.
     
  4. Oct 13, 2007 #3

    morphism

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    Why is |GL(2, Z_3)| odd? Isn't it just 48? In fact, the order of GL(2, Z_p) is even for all primes p.

    But anyway, why would you think that GL and the dihedral group are isomorphic? The latter is much nicer, e.g. it has a presentation with 2 generators.

    Ironically, a proof that GL(2, Z_p) isn't going to be isomorphic to the dihedral group having the same order follows from the fact that of how simple the center (which I believe you're calling the "commutant") of D_n is: it's either isomorphic to Z_1 or Z_2 (the cyclic groups of order 1 or 2, resp.), depending on whether n is odd or even, resp., while the center of GL(2, Z_p) is bigger when p>3. (This doesn't rule out the case when p=3, however; it turns out |C(GL(2, Z_3))| = |C(D_24)| = 2.)

    So I suggest you abandon this particular approach. It's not hard to "just do it": what matrices commute with all the other ones?
     
    Last edited: Oct 13, 2007
  5. Oct 13, 2007 #4
    It was a wild passionate fantasy of the moment:wink: I didn't really take the time to think about it before briging it up. The definition for commutant of a group G that we are using in class is the set of all [itex]xyx^{-1}y^{-1}[/itex] with for all x, y in G. But I actually solved it. Really simple! The det of any any element in the commutator of GL is 1, so the element is in SL. But any element in SL is invertible, and hence can be written as a commutator of two invertible matrices in GL. Hence the commutant of GL(2,Z_p) is SL(2,Z_p).
     
    Last edited: Oct 13, 2007
  6. Oct 13, 2007 #5

    morphism

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    I just saw your post in the Homework forum (isn't it against the rules to post the same thread across forums?), and it turns out you mean "commutator subgroup" when you say "commutant", and not "center".

    And you did get it right: the commutator subgroup of GL(2, Z_p) is SL(2, Z_p). Well, almost. If p=2, then since the GL(2, Z_2) =~ D_3, it follows that the comm subgrp of GL in this case is that of D_3, which is (isomorphic to) Z_3, while |SL(2, Z_2)| = 6.

    Also, I'm not very convinced by your proof of the converse, i.e. that every element of SL(2, Z_p) arises as a commutator of GL(2, Z_p). You said that: "Also any X in SL(n = 2, Z_p) is invertible and hence can be written as a commutator of two invertible matrices over Z_p." But doesn't this apply to every X in GL as well? Thus we're lead to conclude that the commutator subgroup of GL is GL itself...
     
  7. Oct 13, 2007 #6
    oops, didn't know this. Sorry to the moderators :)

    Nice. But you mean it works for all other p?


    You are totally right, that was care free. I am thinking of writing X as a commutator of two matrices with det 1 which would seem to do it. But I don't know how to get around the p = 2 singularity (maybe even 3).
     
  8. Oct 13, 2007 #7

    morphism

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    Yeah - for all odd primes, the commutator subgroup of GL(2, Z_p) is SL(2, Z_p).
     
  9. Oct 13, 2007 #8
  10. Oct 13, 2007 #9

    morphism

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    I don't know of any easy proof. Try the case GL(2, Z_3) - it might be instructive. Or go speak with your professor to see what he/she has in mind.
     
  11. Oct 13, 2007 #10
    Ye, my prof gave that question in class, and I really want to see what he has to say about that. Thank you.
     
  12. Oct 14, 2007 #11
    Hmm, I don't think it can apply to any X in GL because not every X in GL has det 1, while any commutator does.
     
  13. Oct 14, 2007 #12

    morphism

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    All you said was that X is invertible, and then you deduced that it can be written as a commutator of two invertible matrices. You didn't mention/use the fact that X has det 1 here..
     
  14. Oct 14, 2007 #13
    Yes I realize that. I was just trying to correct myself. My problem with my "proof" is that it seems from it that it works for any integer p, where we know this is not true. Hence there must be a fundamental point I am missing. Is the actual proof that complicated to not even dream of reading it this year? If you know any link with that... Thanks.
     
  15. Oct 16, 2007 #14

    morphism

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    I don't know! At any rate, the proof I have in mind is tedious.

    Did you speak to your professor?
     
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