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Centerless groups

  1. Feb 20, 2014 #1
    I imagine a matrix group, with multiplication as the composition rule, to always possess the quality of having centre (I,-I), as I can't see when both elements wouldn't commute with all others. On the other hand, though, a centerless group is defined as having trivial centre, i.e. Z=I (which means, Z doesn't include -I).

    I imagine non-matrix groups could show this property, but I can't think of any.

    Could somebody give a couple of examples of centreless groups, and what "constraints" must be relaxed (from my matrix group example above) in order to achieve them?
     
  2. jcsd
  3. Feb 20, 2014 #2

    pasmith

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    A matrix group must contain the identity [itex]I[/itex], but need not contain [itex]-I[/itex].

    The group [itex]D_3[/itex], which is the symmetry group of an equilateral triangle, has trivial center. The group has a 2-dimensional representation generated by a reflection in the x-axis
    [tex]
    M_m = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}
    [/tex]
    and a rotation through [itex]2\pi/3[/itex] about the origin,
    [tex]
    M_\rho = \begin{pmatrix} \cos(2\pi/3) & -\sin(2\pi/3) \\ \sin(2\pi/3) & \cos(2\pi/3) \end{pmatrix}
    [/tex]
     
  4. Feb 20, 2014 #3
    Thanks, I see how it would work for D3. However, I was referring to the center of SU(2), which is (I,-I).

    If the composition rule is multiplication, how is it possible to find an element of the group which doesn't commute with -I?
     
  5. Feb 20, 2014 #4

    pasmith

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    SU(2) is a matrix group. There are other groups of 2x2 matrices, and the fact that the center of SU(2) is {I, -I} has no bearing on the center of any other matrix group.

    It isn't, but that doesn't matter if the group in question doesn't contain -I in the first place!

    The center of a group [itex]G[/itex] consists exactly of those [itex]g \in G[/itex] such that for all [itex]h \in G[/itex], [itex]gh = hg[/itex].

    Thus, if [itex]G[/itex] is a matrix group and [itex]-I \notin G[/itex], we don't care that [itex](-I)M = M(-I)[/itex] for all [itex]M \in G[/itex]; [itex]-I[/itex] fails to be in the center of [itex]G[/itex] by virtue of not being in [itex]G[/itex] in the first place.
     
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