[proofplan] We use the algebraic definition of the Littlewood-Richardson coefficients as the structure constants for multiplication in the Schur basis of the symmetric-function ring. Since $\Lambda$ is a commutative ring, the products $s_\lambda s_\mu$ and $s_\mu s_\lambda$ are equal. Expanding both products in the Schur basis and comparing the coefficient of $s_\nu$ gives the desired equality. [/proofplan] [step:Expand both products in the Schur basis] Let $\mathcal{P}$ denote the set of all partitions. Since $\{s_\rho : \rho \in \mathcal{P}\}$ is a $\mathbb{Z}$-basis of $\Lambda$, every element of $\Lambda$ has a unique finite expansion in this basis. By the definition of the Littlewood-Richardson coefficients, the products $s_\lambda s_\mu$ and $s_\mu s_\lambda$ have expansions \begin{align*} s_\lambda s_\mu &= \sum_{\rho \in \mathcal{P}} c^\rho_{\lambda,\mu} s_\rho, \\ s_\mu s_\lambda &= \sum_{\rho \in \mathcal{P}} c^\rho_{\mu,\lambda} s_\rho. \end{align*} Here only finitely many coefficients in each sum are nonzero, because multiplication in $\Lambda$ produces a well-defined symmetric function. [/step] [step:Use commutativity of multiplication in $\Lambda$] The ring $\Lambda$ is commutative, so multiplication of symmetric functions satisfies \begin{align*} s_\lambda s_\mu = s_\mu s_\lambda. \end{align*} Substituting the two Schur-basis expansions from the previous step gives \begin{align*} \sum_{\rho \in \mathcal{P}} c^\rho_{\lambda,\mu} s_\rho = \sum_{\rho \in \mathcal{P}} c^\rho_{\mu,\lambda} s_\rho. \end{align*} [guided] The point of passing to the ring $\Lambda$ is that the Littlewood-Richardson coefficients are exactly the coordinates of a product in the Schur basis. The hypotheses give partitions $\lambda$ and $\mu$, hence Schur functions $s_\lambda$ and $s_\mu$ in $\Lambda$. Because $\Lambda$ is a commutative ring, their product does not depend on the order: \begin{align*} s_\lambda s_\mu = s_\mu s_\lambda. \end{align*} Now expand each side in the Schur basis. By definition of $c^\rho_{\lambda,\mu}$ and $c^\rho_{\mu,\lambda}$, we have \begin{align*} s_\lambda s_\mu &= \sum_{\rho \in \mathcal{P}} c^\rho_{\lambda,\mu} s_\rho, \\ s_\mu s_\lambda &= \sum_{\rho \in \mathcal{P}} c^\rho_{\mu,\lambda} s_\rho. \end{align*} Since the two products are equal as elements of $\Lambda$, these two Schur-basis expansions represent the same element: \begin{align*} \sum_{\rho \in \mathcal{P}} c^\rho_{\lambda,\mu} s_\rho = \sum_{\rho \in \mathcal{P}} c^\rho_{\mu,\lambda} s_\rho. \end{align*} This is the exact place where commutativity is used: it turns equality of the two ordered products into equality of their Schur-coordinate expansions. [/guided] [/step] [step:Compare the coefficient of $s_\nu$] Since $\{s_\rho : \rho \in \mathcal{P}\}$ is a basis of $\Lambda$, coordinates in this basis are unique. Therefore, from \begin{align*} \sum_{\rho \in \mathcal{P}} c^\rho_{\lambda,\mu} s_\rho = \sum_{\rho \in \mathcal{P}} c^\rho_{\mu,\lambda} s_\rho, \end{align*} the coefficients of the basis vector $s_\nu$ must agree. Hence \begin{align*} c^\nu_{\lambda,\mu} = c^\nu_{\mu,\lambda}. \end{align*} This proves the commutativity symmetry for all partitions $\lambda$, $\mu$, and $\nu$. [/step]