Question:

Suppose that $V_1$ and $V_2$ are two finite dimensional representations of lie algebra $\mathfrak{g}$ generated by highest weight vectors $v_1^*$ and $v_2^*$ of weights $\mu_1$ and $\mu_2$ respectively.

Now, suppose that $V_3$ is an irreducible representation of $V_1 \otimes V_2$ such that the highest weight of $V_3$ has weight $\mu$. I want to show that $\mu = \mu_1 + \lambda $ for some weight $\lambda$ of $V_2$.

Approach: I know that the weight spaces of the tensor $V_1 \otimes V_2$ are spanned by tensor products of weight vectors of $V_1$ and $V_2$. Then $v_1 \otimes v_2$ is a highest weight vector of $V_1 \otimes V_2$ of highest weight $\mu_1+ \mu_2$. If $w$ is a highest weight vector of $V_3$, then it is a sum of the form $v_{1,k} \otimes v_{2,k}$ with $v_{1,k}$ of weight $\lambda_{1,k}$ and $v_{2,k}$ of weight $\lambda_{2,k}$ such that $\lambda_{1,k} + \lambda_{1,k} =\mu.$

The goal is to show that for one of these tensors $\lambda_{1,k} = \mu_1.$ Considering a simpler case when $w = v_1 \otimes v_2$ with $v_1$ of weight $\lambda_1$, $v_2$ of weight $\lambda_2$ and $\lambda_1 +\lambda_2 = \mu$. Suppose that $\lambda_1 \neq \mu_1$, how can I get a contradiction here? I believe if this part is cleared, we can then next suppose that $w = \sum_{k}v_{1,k} \otimes v_{2,k}$ and also that $\lambda_{1,k} \neq \mu_1$ for all $k$ ( the sum being over some indexing set) and obtain a contradiction ( I believe this approach answers this question).

P.S: I apologize this is not a research question but I think its an interesting question. I have been struggling to come up with a proof and that is why I am seeking help. I will be glad if anyone interested in this question to please provide me a proof/explanation to this question. Thanks