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A certain finite group is of order 10

  1. Mar 28, 2005 #1
    Hi, this problem has been giving me difficulty. A certain finite group is of order 10. Does it have a 3-dimensional irreducible representation? Why or why not? Does anyone know?
  2. jcsd
  3. Mar 28, 2005 #2

    matt grime

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    There are several results that imply it cannot have an irreducilbe 3-d representation (over C). The most far reaching is that the order of every simple (irreducible) representation divides the order of the group. This is a hard theorem to prove, so yuo should avoid relying on it, though it is very powerful.

    A simpler result, and an obvious one if you know a little character theory, is that the sum of the squares of the irreducible representations' dimensions is the order of the group.

    Now, this group has at least three conjugacy classes, right, since the identity, some element of order 2 and some element of order 5 all exist and are not conjugate, so there are at least 3 irred reps.

    If we suppose there is a 3-d one, plus the obvious trivial rep, then 1+9=10 already, so there can't be a thrid one which we know must exist. Thus there is no irred rep of dimension 3.
  4. Mar 28, 2005 #3
    But how do you know what the conjugacy classes are, and what do you mean they are not conjugate? So 1^2 + 3^2+3^2 = 10, which is the order of the group, and tells me I have 3 irreducible reps, but what is meant by 3-dimensional? Thanks a lot
    Last edited: Mar 28, 2005
  5. Mar 29, 2005 #4

    matt grime

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    1+3^2+3^2=19, not 10.

    There are at least 3 conjugacy classes. You know what a conjugacy class is? Elements in the same conjugacy class have the same order, thus the elements in the sylow 2 subgrops and the sylow 5 subgroups cannot be conjugate, and only the identity is conjugate to the identity, thus there are at least 3 conjugacy classes, and hence at least 3 irreducible representations.

    3 dimensional refers the the representation.
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