Character table for cyclic group of order 7

[CAF123]

Homework Statement

a)Write down all irreducible representations of $\mathbb{Z}_7$.
b)How many of the irreducible representations are faithful?

Homework Equations

Group structure of $\mathbb{Z}_7 = \left\{e^{2\pi in/7}, \cdot \right\}$ for $n \in \left\{0,...,6\right\}$

The Attempt at a Solution

I am trying to construct the character table of $\mathbb{Z}_7$. Since this is an abelian group, there are 7 conjugacy classes and thus 7 one dimensional irreducible representations. Since the representations are all 1D, the character of the representation is simply the representation itself. My question is are the representations of the form $e^{2\pi in/7}?$ Letting n run from 0 to 6 does indeed give 7 irreducible representations.

I have in some notes the following statement: Let $\rho$ denote a representation. Then $$\rho(1)^7 = \rho(1^7) = \rho(1) = 1$$ Since $\rho$ is always 1D, it can be written as a complex scalar. Let $x = \rho(1) \Rightarrow x^7 - 1 = 0$ and thus the seven irreducible reps of 1 are of the form $e^{2\pi in/7}$ Do I do a similar analysis for the other elements? I wasn't sure though why $\rho(1^7) = \rho(1)$ though. $1 \in \mathbb{Z}_7 = \left\{n, + | n \in \left\{0,...,6\right\}\right\}$, so I think $1^7 = 1 + 1...+1 = 0$ mod$7$, by summing 1 seven times.

To mentors: This is from a course on symmetries of QM, so that is why I posted it here. Perhaps it is better elsewhere.

Thanks.

 

[mighty2000]

Hello! Great question. You are correct in thinking that the irreducible representations of $\mathbb{Z}_7$ are of the form $e^{2\pi in/7}$ for $n \in \left\{0,...,6\right\}$. This is because the group structure of $\mathbb{Z}_7$ is isomorphic to the group of rotations in the complex plane, and so the irreducible representations are just the characters of these rotations.

In terms of your question about why $\rho(1^7) = \rho(1)$, this is because the representation of $\mathbb{Z}_7$ is defined by mapping each element of the group to a matrix. So, the representation of the identity element (which is just the number 1) is just the identity matrix, which is the same as the representation of any power of 1. This is because the identity element is always mapped to the identity matrix under any representation.

As for the second part of the question, there are 7 irreducible representations of $\mathbb{Z}_7$ and all of them are faithful. This is because the group is abelian, meaning that all elements commute with each other, and so the kernel of any representation must be trivial (i.e. only the identity element is mapped to the identity matrix). Therefore, all of the irreducible representations are injective and thus faithful.

I hope this helps! Let me know if you have any further questions.