# Trivial representation from the Young symmetrizer

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.

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 $$\alpha=\sum_{g\in S_d} \mu_g\in\mathbb C$$.