I have this exercise that I can't do.
Let $g$ be a solvable lie algebra. Show that the set of isomorphisms classes os irreducible representations of $g$ is in bijection with $(g/g')^*$ where $g'=[g,g]$ and the * is for the dual space.
I did show that for the solvable lie algebras every irreducible representation have dimension $1$ but don't know how to continue. I was trying to get a morphism between $g$ and that set that was surjective and that $g'$ was the kernel for that morphism, then I would have a isomorphism with $g/g'$ and because I'm working with lie algebras of finite dimension that space is equal to its dual, but I can't find that morphism. Can someone help? Is there a better way of doing this?
If you've already shown that every irreducible representation $V$ is one-dimensional, that means up to isomorphism it is of the form $\mathfrak g \rightarrow F$ ($F$ your ground field, presumably $\mathbb C$), and $\mathfrak g'$ is in its kernel. So that gives you one way, although you might want to check how much this depends on the chosen isomorphism of reps $V\simeq F$. A map the other way around is even easier to get. Now show they compose to the identity.