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Let $z$ be a generator of the cyclic group $\mathbb{Z}_3 = \{ 1,z,z^2 \}$. Prove that a representation $\rho$ of $\mathbb{Z}_3$ in the $2$-dimensional complex vector space $\mathbb{C}^2$ can be defined by $$\rho(z) = \begin{bmatrix} -1 & 1 \\ -1 & 0 \end{bmatrix}.$$

I know I need to check that $\rho : \mathbb{Z}_3 \to GL(\mathbb{C}^2)$ is a homomorphism, but how do I check that $\rho(g \circ h) = \rho(g) \circ \rho(h)$ for any $g,h \in \mathbb{Z}_3$?

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Since any $g,h$ can be written as $g=z^a.h=z^b$ we have $\rho(gh)=\rho(z^{a+b}).$ Implicit in the question is that $\rho(z^c):=A^c$ for any $c=0,1,2,$ where $A$ is the given matrix. Now you know that $\rho(gh)=A^{a+b}=A^aA^b=\rho(g)\rho(h).$

The important property that you need to check is that $A^3=I$ is the identity matrix.

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First thing to do is make sure that $\rho(z)=\left(\begin{array}{cc}-1&1\\-1&0\end{array}\right)$ is really an element of $GL_2(\mathbb{C})$. It's obvious that $\rho(z)\in M_{2\times 2}(\mathbb{C})$, but you should prove it's invertible.

Then, just write it out. $\mathbb{Z}_3$ is cyclic, so the entire representation is determined by $\rho(z)$. If you can prove that $\rho(z)$ has order $3$ in $GL_2(\mathbb{C})$ you're done. So, calculate $\rho(z)^2$ and $\rho(z)^3$ and make sure the latter is the identity matrix and the former is not.

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