Representations of the cyclic group of order n

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SUMMARY

The discussion centers on the representation of the cyclic group of order m, denoted as G = < a : a^m = 1 >, using a matrix A in GL(n, ℂ). It establishes that a mapping ρ : G → GL(n, ℂ) defined by ρ(a^r) = A^r is a representation if and only if A^m = I. Participants confirm that multiple matrices A satisfying A^m = I yield different representations, emphasizing that the mapping ρ itself is the representation, not the matrix A.

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  • Understanding of cyclic groups and their properties
  • Familiarity with group representations and homomorphisms
  • Knowledge of the general linear group GL(n, ℂ)
  • Basic linear algebra concepts, particularly matrix exponentiation
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  • Explore the concept of group homomorphisms in detail
  • Study the structure and properties of GL(n, ℂ)
  • Learn about the classification of representations of finite groups
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Math Amateur
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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.

===================================================================

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

Can someone please confirm this?
 
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Math Amateur said:
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.

===================================================================

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

Can someone please confirm this?

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.
 
Thanks

Just needed the confirmation before going onwards!

Thanks again
 

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