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In Proposition 2.4.5 of Sagan's book The Symmetric Group, he proves that if $\operatorname{Hom}(S^\lambda , M^\mu)$ is nonzero, then $\lambda\ge \mu$ where $S^\lambda$ is a Specht module and $M^\mu$ is a module of tabloids. I don't understand this line:

$$\theta(\kappa_t\{t\}) = \kappa_t \theta(\{t\})$$

I can't see any reason why $\theta$ and $\kappa_t$ should commute.

Notation: $\theta$ is some arbitrary (nonzero) element of $\operatorname{Hom}(S^\lambda, M^\mu)$, $\kappa_t\{t\}$ for a tableau $t$ is $\sum_{\pi\in C_t}(\operatorname{sgn} \pi)\{\pi t\}$.

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The map $\theta$ is an $S_n$-module homomorphism, so $\theta$ is linear and $\theta(\sigma v)=\sigma\theta(v)$ for all $\sigma\in S_n$ and all $v\in S^\lambda$. Since $\kappa_t\in\mathbb{C}S_n$, these two properties of $\theta$ imply $$\theta(\kappa_t\{t\})=\kappa_t\theta(\{t\}).$$

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  • $\begingroup$ Ah, looks like I forgot the definition of Hom. Thanks for pointing this out. $\endgroup$ – Cyclicduck Sep 4 '18 at 18:38

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