# Representations of the cyclic group of order n

1. Jan 21, 2012

### Math Amateur

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 : $a^m = 1$ >.

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

$\rho : a^r \rightarrow A^r \ \ (0 \leq r \leq m-1 )$

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

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

Suppose $\rho$ is a representation of G. Then

$I = 1 \rho = ( a^m ) \rho = {(a \rho)}^m = A^m$

Conversely assume that $A^m = I$. Then $( a^i ) \rho = A^i$ for all integers i.

Therefore for all integers i, j

$( a^i a^j ) \rho \ = \ ( a^{i+j} ) \rho \ = \ A^{i+j} \ = \ A^i A^j \ = \ ( a^i \rho ) a^j \rho )$

and so $\rho$ is a representation.

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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 $A^m = I$

2. Jan 21, 2012

### Deveno

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.

3. Jan 21, 2012

### Math Amateur

Thanks

Just needed the confirmation before going onwards!

Thanks again