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Center and Commutant: GL(n = 2, Z_p)

  1. Oct 13, 2007 #1
    1. The problem statement, all variables and given/known data

    What is the center and the commutant of GL(n = 2, Z_p)?

    2. Relevant equations

    The center of a group G: z(G) = {x in G| xy = yx for all y in G}

    Commutant of a group G is the set of all [itex] xyx^{-1}y^{-1} [/itex] where all x, y are in G.

    3. The attempt at a solution

    I tried some calculations with p = 2 first to see if some pattern emerged, but gave up along the way multiplying matrices. My idea is that since I can get the order of GL(n = 2, Z_p) for any p, then I could try to construct an isomorphism between this group and the dihedral group that has the same order. Since the center and commutant of this dihedral group is much easier to get, then I will have obtained the center and commutant of the GL group. Is this right? I gave a try to p=2 that gives order 6 to GL and was able to construct an isomorphism successfully with D_3. My problem now is trying to prove that I can show the isomorphism between any GL(n = 2, Z_p) and the corresponding dihedral group with the same order. Thanks for any help.
  2. jcsd
  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
    Oh I just realized how simple it is. Take any A, B in GL(n = 2, Z_p). Then

    [itex]det(ABA^{-1}B^{-1}) = 1 [/itex]

    the commutant of GL(n = 2, Z_p) is a subset of SL(n = 2, Z_p).

    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. Hence the commutant of GL(n = 2, Z_p) is SL(n = 2, Z_p).
    Last edited: Oct 13, 2007
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