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A decomposition of tensor product of representations of $\mathrm{GL}_l(\mathbb{C})$ and a decomposition of Littlewood-Richardson numbers?

For a positive integer $m$, denote by $T(m)=\{(\lambda_1,\dots,\lambda_m)\in \mathbb{Z}^m:\lambda_1\ge \lambda_2\ge\dots \ge\lambda_m\}$ and $T^+(m)=\{ (\lambda_1,\dots,\lambda_m)\in \mathbb{Z}^m:\lambda_1\ge \lambda_2\ge\dots \ge\lambda_m\ge 0\}$. For $\lambda=(\lambda_1,\dots,\lambda_m)\in T^+(m)$, denote by $|\lambda|=\lambda_1+\dots+\lambda_m$. Then an element $\lambda$ in $T^+(m)$ can be viewed as a partition of $|\lambda|$. For an element $\lambda\in T(m)$, denote by $\Psi_\lambda$ the irreducible representation of $\mathrm{GL}_m(\mathbb{C})$ with maximal weight $\lambda$. Note that if $\lambda=(\lambda_1,\dots,\lambda_m)$, then $\lambda^+:=(\lambda_1-\lambda_m,\lambda_2-\lambda_m,\dots,\lambda_{m-1}-\lambda_m,0)\in T^+(m)$ and $\Psi_{\lambda}=\Psi_{\lambda^+}\otimes D_{\lambda_m}$, where $D$ denotes the determinant and $D_{\lambda_m}=\det^{\lambda_m}$. Moreover, if we denote $\lambda^*=(-\lambda_m,\dots,-\lambda_1)$, then it is known that the dual of $\Psi_{\lambda}$ is $\Psi_{\lambda^*}$. Notations and reference is Fulton-Harris' book "Representation theory: a first course".

Now let $l,m,n$ be positive integers such that $m+n<l$ and $n\le m$. Let $\lambda\in T^+(m),\mu\in T^+(n)$ and we consider the decomposition $$\Psi_{(\lambda,0_{l-m})}\otimes \Psi_{(0_{l-n},\mu^*)}$$ as a representation of $\mathrm{GL}_l(\mathbb{C})$.

Question: Is the following decomposition true $$\Psi_{(\lambda,0_{l-m})}\otimes \Psi_{(0_{l-n},\mu^*)}=\bigoplus_{x\in T^+(m),y,z\in T^+(n)}N_{xz\lambda}N_{yz\mu}\Psi_{(x,0,y^*)}?$$ Here $N_{xz\lambda}$ is the Littlewood-Richardson numbers, which is usually written as $c_{xz}^\lambda$ in other literatures.

Note that, we can write $\Psi_{(0_{l-n},\mu^*)}=\Psi_{\tilde\mu}\otimes D_{-\mu_1}$ with $\widetilde \mu=(\mu_1,\dots,\mu_1,\mu_1-\mu_n,\dots,\mu_1-\mu_2,0)$ with $(l-n)$'s $\mu_1$, then by Littlewood-Richardson rule, we can decompose the left side as \begin{align*} \Psi_{(\lambda,0_{l-m})}\otimes \Psi_{(0_{l-n},\mu^*)}&=\Psi_{(\lambda,0_{l-m})}\otimes \Psi_{\tilde\mu}\otimes D_{-\mu_1}\\ &=\bigoplus_{\nu}N_{\tilde \mu \lambda \nu}\Psi_{\nu}\otimes D_{-v_1} . \end{align*} It is not hard to see that if $N_{\tilde \mu \lambda \nu}\ne 0$, then $\nu$ must be of the form $(\nu_1,\dots,\nu_l)$ with $\nu_{m+1}=\dots=\nu_{l-n-1}=v_1$, which means that $ \Psi_{\nu}\otimes D_{-v_1} $ is of the form $\Psi_{(x,0,y^*)}$ with $x\in T^+(m)$ and $T^+(n)$. Thus the above question is the same as the following

Question': Is the following identity true $$N_{\tilde \mu \lambda \nu}=\sum_{z\in T^+(n)}N_{xz\lambda}N_{yz\mu},$$ for any $\lambda,x\in T^+(m),\mu,y\in T^+(n)$ (one can add the condition $|\lambda|-|\mu|=|x|-|y|$)? Here $\tilde \mu=(\mu_1,\dots,\mu_1,\mu_1-\mu_n,\dots,\mu_1-\mu_2,0)$, and $\nu=(x,0,y^*)+(v_1,\dots,v_1).$

I checked several small examples and it seems that the above questions have positive answer. I believe they should be true in general. I am wondering if this is a well-known result. Any comments and references will be appreciated. Thanks in advance.

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