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Let $U$ be a submodule of the Young permutation module $M^{\lambda }$ over any field. Then eitherfor the tabloid bilinear form. The key identity is $b_{t}u=⟨u,e(t)⟩e(t)$: if one pairing is nonzero, the cyclic generator $e(t)$ and hence the whole Specht module lies in $U$.

The James submodule theorem says that for every $F{S}_n$-submodule $U\le M^{\lambda }$, eitherwhere orthogonality is taken with respect to the tabloid bilinear form.

Fix a $\lambda$-tableau $t$. Part a shows that for every tabloid $\{s\}$, the vector $b_t\{s\}$ is either zero or a signed copy of the polytabloid $e(t)$. Comparing the coefficient of $\{t\}$ gives the precise identityIf $U⊈(S^{\lambda }{)}^{\perp }$, choose $u\in U$ and a tableau $t$ with $⟨u,e(t)⟩\ne 0$. Since $U$ is a submodule, the identity puts $e(t)$ in $U$. Every polytabloid of shape $\lambda$ is an $S_n$-translate of $e(t)$, so their span $S^{\lambda }$ lies in $U$. If no such $u,t$ exist, then by definition $U\subseteq (S^{\lambda }{)}^{\perp }$. This proves the theorem over the arbitrary field $F$.

Because $C(t^{\prime })=R(t)$,Applying $\theta$ and using $re(u)=\mathrm{sgn}(r)e(u)$ givesThus $m={\sum }_{r\in R(t)}re(t)$. Every $r$ fixes the tabloid $\{t\}$, while $e(t)$ has coefficient one at $\{t\}$. Invariance of the tabloid bilinear form now yields

The assumed one-dimensional-image property givesfor some $c\in \mathbb{F}$. The coefficient of $\{t\}$ in $e(t)$ is one, soThe tabloid bilinear form is invariant, and the involution on the group algebra fixes the Column antisymmetrizer of a Young tableau because inversion preserves sign. Thereforewhich proves the formula.

If some $a_j\ge p$, then $a_j!=0$ in $\mathbb{F}$. Part b(ii) makes every pairing $⟨e(t),e(u)⟩$ zero, so the Tabloid bilinear form vanishes identically on $S^{\lambda }$ and $S^{\lambda }\subseteq (S^{\lambda }{)}^{\perp }$.

Conversely, if $\lambda$ is $p$-regular, every $a_j!$ is nonzero. Fact 1 supplies tableaux $t,u$ withThus the restriction of the form is not identically zero and $S^{\lambda }⊈(S^{\lambda }{)}^{\perp }$.