Question about Schur's lemma and irreducible representations of $S_n$

Schur's lemma says that if $M,N$ are two irreducible representations of a group $G$, then either $Hom_G(M,N)=0$ if $M,N$ are not isomorphic, or every $\varphi\in Hom_G(M,N)$ is invertible if they are isomorphic.

If we look at the case $G=S_n$ and $M=V_\lambda,\ N=V_\mu$ for partitions $\lambda,\mu$ of $n$, we must have $Hom_{S_n}(V_\lambda,V_\mu)=0$ if $\lambda\neq \mu$. But $V_\lambda$ is self dual, so $Hom_{S_n}(V_\lambda,V_\mu)\cong V^*_\lambda\otimes V_\mu\cong V_\lambda\otimes V_\mu=0$.

But exercise 4.51 in Fulton-Harris implies that the tensor product $V_\lambda \otimes V_\mu$ decomposes as a direct sum of copies of $V_v$ for partitions $v$ of $n$. In particular, $V_{(n)}\otimes V_\lambda=V_\lambda$ and $V_{(1,...,1)}\otimes V_\lambda=V_{\lambda^\prime}$, where $\lambda^\prime$ is the conjugate partition.

What went wrong here?

You're confusing the internal and external hom here. $V_{\lambda}^{\ast} \otimes V_{\mu}$ is isomorphic to the internal hom $[V_{\lambda}, V_{\mu}]$, which is the vector space of all linear maps $V_{\lambda} \to V_{\mu}$ with the appropriate $S_n$-action. The external hom $\text{Hom}_{S_n}(V_{\lambda}, V_{\mu})$ is the $S_n$-invariant subspace of the internal hom.
This implies that $V_{\lambda} \otimes V_{\mu}$ has at most one trivial summand (it has one if $\lambda = \mu$ and zero otherwise), but it can have lots of other nontrivial summands.
• I guess, but if the OP knew that there even was such a thing as the external tensor product he would hopefully be denoting it by $\otimes_{S_n}$. – Qiaochu Yuan Dec 13 '15 at 20:41