Let $G$ be a finite group, $|G|=n$, and $g,h\in G$ such that $gh=hg$. If $\rho: G \rightarrow GL(V)$ is a representation, where $V$ a vector space over $\mathbb{C}$, prove that there exists a basis of $V$ in which both $\rho(g)$ and $\rho(h)$ are diagonal.
I could prove this result using general methods of linear algebra, as follows:
Since $G$ is finite, $|G|=n$, it follows that $\rho(g)^n =\rho(g^n) = \rho(1) = I$. If $p(X) = X^n-1,$ the condition before shows us that $p(\rho(g))=0$, which ensures that the minimal polynomial for $\rho(g)$ must divide $X^n-1$. Now, $X^n-1$ has no repeated roots and therefore so does the minimal polynomial. We conclude from this that $\rho(g)$ is diagonalizable.
The same argument runs for $\rho(h)$. Now, both are diagonalizable and commuting. By linear algebra, they are simultaneously diagonalizable.
My question is: is there a way to prove this claim using only representation theory methods? or a proof which is heavily based on representation theory methods?
Since $G$ is abelian, then every irreducible representation is 1-dimensional. By Maschke's Theorem ($G$ is finite), every finite-dimensional representation is semisimple, hence every representation of $G$ is a direct sum of 1-dimensional representations $V=\mathbb{C}v_1\oplus...\oplus \mathbb{C}v_n$. It follows $\{v_1,...,v_n\}$ is a common basis of $\rho(g)$ for all $g\in G$.