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Representations of the cyclic group of order n

  1. Jan 21, 2012 #1
    I am reading James and Liebeck's book on Representations and Characters of Groups.

    Exercise 1 of Chapter 3 reads as follows:

    Let G be the cyclic group of order m, say G = < a : [itex] a^m = 1 [/itex] >.

    Suppose that A [itex] \in GL(n \mathbb{C} ) [/itex], and define [itex] \rho : G \rightarrow GL(n \mathbb{C} ) [/itex] by

    [itex] \rho : a^r \rightarrow A^r \ \ (0 \leq r \leq m-1 ) [/itex]

    Show that [itex] \rho [/itex] is a representation of G over [itex] \mathbb{C} [/itex] iff and only if [itex] A^m = I [/itex]

    The solution given is as follows (functions are applied from the right)

    Suppose [itex] \rho [/itex] is a representation of G. Then

    [itex] I = 1 \rho = ( a^m ) \rho = {(a \rho)}^m = A^m [/itex]

    Conversely assume that [itex] A^m = I [/itex]. Then [itex] ( a^i ) \rho = A^i [/itex] for all integers i.

    Therefore for all integers i, j

    [itex] ( a^i a^j ) \rho \ = \ ( a^{i+j} ) \rho \ = \ A^{i+j} \ = \ A^i A^j \ = \ ( a^i \rho ) a^j \rho ) [/itex]

    and so [itex] \rho [/itex] is a representation.


    That is all fine but I am just getting into representations and wish to get an intuitive understanding of what is happening.

    In the above I am suprised that we have no explicit form for A.

    Does this mean we have many representations for G in this case - that is any and every matrix A for which [itex] A^m = I [/itex]

    Can someone please confirm this?
  2. jcsd
  3. Jan 21, 2012 #2


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    Science Advisor

    well the matrix A isn't the representation, the mapping ρ is. and this mapping doesn't have to be an isomorphism, just a homomorphism. for example, ρ could be trivial, sending every element of G to the identity matrix. or, if m is even, ρ could send a generator of Cm to -I. more generally, with an nxn matrix, there can be many more m-th roots of the identity matrix than you can have in a field. so, yes, representations aren't unique.
  4. Jan 21, 2012 #3

    Just needed the confirmation before going onwards!

    Thanks again
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