Let $C_n$ be the cyclic group order n. Let V be the faithful two dimensional representation (over complex field) denoted by:
\begin{equation}
\rho(g^j)=
\begin{pmatrix} \omega^j & 0 \\ 0 & \omega ^{-j} \end{pmatrix} \end{equation}
Where $\omega^j = $exp$(\frac{2\pi i}{n}j)$ and $j=1,\dots ,n$.
What are the irreducibles of $C_n$ in the direct sum decomposition of $V\otimes V$?
My attempt:
$C_n$ is cyclic (hence abelian) thus irreducible representations are one dimensional. There are n irreducible representations $V_h$ corresponding to the character $\chi_h(g) = \omega ^h$.
If $V_h$ apprears in the decomposition of $V \otimes V$ then the inner product $<\chi, \chi_h>$ is nonzero.
Now \begin{equation}\chi(g^j)=\chi_{V\otimes V}(g)= \omega^{2j}+\omega^{-2j} +2\end{equation}
Thus \begin{equation} <\chi,\chi_h>= \frac{1}{n}\Sigma_{j=1}^{n}(\omega^{j(2-h)}+\omega^{-j(2+h)}+2\omega^{-jh}) \end{equation}
Now this sum will equal 0 (by geometric series formula) unless $h=2,0$.
So $V_2,V_0$ are the only irreducibles appearing in $V \otimes V$.
Is this the correct way of approaching the problem? If not what is?
Label the irreducible representations of $C_n$ by $\{V_k \mid k \in \mathbb{Z} / n \mathbb{Z}\}$, where $V_k$ is a one-dimensional representation where your chosen generator of $C_n$ acts by $\exp(2 \pi i k / n)$. Then it is clear that you have $V \cong V_1 \oplus V_{-1}$ as representations. Since tensor product distributes over direct sum, we must have $$ \begin{aligned} V \otimes V & \cong (V_1 \oplus V_{-1}) \otimes (V_1 \oplus V_{-1}) \\ & \cong (V_1 \otimes V_1) \oplus (V_1 \otimes V_{-1}) \oplus (V_{-1} \otimes V_{1}) \oplus (V_{-1} \otimes V_{-1}) \\ & \cong V_2 \oplus V_0 \oplus V_0 \oplus V_{-2} \end{aligned}$$ where the very last line follows from a direct computation that $V_k \otimes V_r \cong V_{k + r}$.