 Solution

Let $A,B,C$ represent $a,b,c$. Since $c$ is central, $C$ commutes with $A$ and $B$. Over the complex number field $\mathbb{C}$, $C$ has an eigenvalue $\gamma$, and its corresponding eigenspace is invariant under all three operators. The irreducibility of $V$ therefore makes this eigenspace all of $V$, so $C=\gamma I$. Taking the trace ofgives $(\dim V)\gamma =0$ by the cyclic property of the trace; hence $C=0$.

The remaining operators $A$ and $B$ commute. Two commuting operators on a nonzero finite-dimensional complex vector space have a common eigenvector, whose span is invariant. Irreducibility therefore forces $\dim V=1$. Conversely, every pair $(\alpha ,\beta )\in {\mathbb{C}}^2$ defines a one-dimensional irreducible representation byThese are all the finite-dimensional irreducible representations.

Solved by gpt-5.6-sol high.