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I am following the notation about Young symmetrizer introduced in Fulton & Harris and here. I know that for a general $d$, the partition $(d)$ of $d$ gives rise the trivial representation of $S_d$. In a general setting, the irreducible representation $V_\lambda$ corresponding to some partition $\lambda$ of $d$ is given by $$V_\lambda=\mathbb CS_d\cdot c_\lambda, $$ where $c_\lambda$ is the Young symmetrizer, in particular, the trivial symmetrizer is given by $\sum_{\sigma\in S_d}e_\sigma$, so that we should have $$\mathbb CS_d\cdot c_\lambda=\mathbb C \cdot \sum_{\sigma\in S_d}e_\sigma. \tag{$*$}$$ In everything I have found online and textbooks, it is claimed that this is the case, but I have not found any proof and I just cannot see it. I do not know what I am missing, for if $v\in \mathbb CS_d$, then we can write $v=\sum_{\sigma\in S_d}\mu_\sigma e_\sigma$, and by right multiplication by $c_\lambda$ yields $$\sum_{g,h\in S_d}\mu_g e_{g\cdot h}.$$ This should just be a scalar times the diagonal, for $(1)$ to be true, but I cannot see this being true.

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Recall that left multiplication by some $g\in S_d$ is a bijection in $S_d$, thus we can relabel our indices $h\to s=g\cdot h$ and now we have $$\sum_{g,h\in S_d }\mu_g e_{g\cdot h}=\sum_{g,s\in S_d} \mu_g e_s=\alpha \sum_{s\in S_d}e_s, $$ where $\alpha=\sum_{g\in S_d} \mu_g\in\mathbb C$.

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