In general, Schur's lemma is stated as follows.
(i) Let $(\pi_1,V_1)$ and $(\pi_2,V_2)$ be irreducible representations, and let $T:V_1\rightarrow V_2$ be an intertwining operator. Then either $T$ is zero or it is an isomorphism.
(ii) Suppose that $(\pi,V)$ is an irreducible representation of $G$ and $T:V\rightarrow V$ is an intertwining operator. Then there exists a scalar $\lambda\in\mathbb{C}$ such that $T(v)=\lambda v$ for all $v\in V$.
Now I'm considering the Lie algebras' version of Schur's lemma in Exercise 10.5 in Lie Groups by Daniel Bump.
Let $\mathfrak{g}$ be a Lie algebra, and let $V$, $W$ be simple $\mathfrak{g}$-modules.
(i') The space of $\mathfrak{g}$-module homomorphisms $\phi:V\rightarrow W$ is at most one-dimensional.
(ii') The space of invariant bilinear forms $V\times W\rightarrow\mathbb{C}$ is at most one-dimensional.
Here are my idea and question to verify Schur's lemma of the version for Lie Algebras:
One can conclude (i') immediately from (i) if there exists an irreducible representation for any Lie algebra (how to demonstrate it or where can I find the result about it?). We may obtain (ii') by a proposition in Bump's Lie Groups:
Proposition 10.5. Let $\mathfrak{g}$ be a finite-dimensional simple Lie algebra over a field $F$. Then there exists, up to scalar, at most one invariant bilinear form on $\mathfrak{g}$. If a nonzere invariant bilinear form exists it is nondegenerate and symmetric.
Can we generalize Proposition 10.5 in the case that $\mathfrak{g}$ is an arbitrary Lie algebra to conclude (ii')? Moreover, what's the relation between the above two versions of Schur's lemma?
Thanks for the comments. We may add some information to the original exercise, that $V$ and $W$ should be simple $\mathfrak{g}$-modules.
For (i'), since simple $\mathfrak{g}$-module is already an irreducible representation, by (i), the space of $\mathfrak{g}$-module homomorphisms is at most one-dimensional.
For (ii'), since a bilinear form $V\times W\rightarrow \mathbb{C}$ gives a $\mathfrak{g}$-linear map from $V$ to the dual of $W$, and it follows that it is an intertwining operator. Then applying (i) we can conclude what we desire.