Use the standard invariant tabloid bilinear form, with the tabloid basis orthonormal. Each permutation acts by an orthogonal matrix, so $⟨gx,y⟩=⟨x,g^{-1}y⟩$. Inversion permutes $C_t$ and preserves its signs. ThereforeThe James submodule theorem states that for any submodule $U\le M^{\lambda }$, either $S^{\lambda }\subseteq U$ or $U\subseteq (S^{\lambda }{)}^{\perp }$. Use the allowed fact ${\kappa }_t{M}^{\lambda }=F{e}_t$. The coefficient of $\{t\}$ in $e_t$ is one, so the displayed adjoint identity gives ${\kappa }_{t}x=⟨x,e_t⟩e_t$. If $U$ is not orthogonal to $S^{\lambda }$, some $x\in U$ and some tableau $t$ have $⟨x,e_t⟩\ne 0$. Hence $e_t\in U$, and its permutation translates span $S^{\lambda }$. This proves the theorem.

Let $R=S^{\lambda }\cap (S^{\lambda }{)}^{\perp }$. Every proper submodule of $S^{\lambda }$ lies in $R$ by the theorem. If the restricted form is nonzero, $R$ is proper, hence is the unique maximal submodule andIn the stronger nondegenerate case specified in the question, $R=0$, so $S^{\lambda }$ itself is simple. The nonzero-form version also covers the possibly degenerate case needed in the final calculation.

For shape $(n-1,1)$, identify $M$ with $F^n$ with basis $v_1,\dots ,v_n$. Its Specht module is the augmentation subspace $S=\{\sum a_i{v}_i:\sum a_i=0\}$, spanned freely by $v_i-v_n$, $1\le i<n$. Its Gram matrix is $G=I_{n-1}+J_{n-1}$, with diagonal entries two and off-diagonal entries one. Equivalently, $S^{\perp }$ in $M$ is the span of $v_1+\cdots +v_n$. This vector belongs to $S$ exactly when $p\mid n$. Thus the restricted radical has dimension one if $p\mid n$ and zero otherwise, givingFor $n\ge 3$ this quotient is nonzero and simple by the theorem. In the small exceptional case $n=2,p=2$, the canonical radical quotient is zero, as the same dimension formula states; it is then not a nonzero simple module. This qualification distinguishes the numerical formula from the nonzero-form hypothesis.

The Killing form is ${\kappa }_L(x,y)=\mathrm{Tr}(\mathrm{ad}x\mathrm{ad}y)$. Write $X=\mathrm{ad}x$, $Y=\mathrm{ad}y$, $Z=\mathrm{ad}z$. Since $\mathrm{ad}[x,y]=[X,Y]$, cyclicity of trace givesIf $x\in I^{\perp }$, $a\in L$ and $y\in I$, invariance gives ${\kappa }_L([a,x],y)={\kappa }_L(x,[y,a])=0$, because $[y,a]\in I$. Thus $I^{\perp }$ is an ideal. For $x,y\in I$, the restrictions of their adjoint maps to $I$ are the adjoint maps of $I$, and their induced maps on $L/I$ vanish. Block traces therefore give ${\kappa }_L{|}_{I\times I}={\kappa }_I$.

If $A$ is an Abelian ideal, $\mathrm{ad}a\mathrm{ad}x$ maps $L$ into $A$ and vanishes on $A$ for $a\in A$. In a basis starting with $A$ it has zero diagonal blocks, hence trace zero. Therefore every Abelian ideal lies in $L^{\perp }$.

Cartan criterion for semisimplicity says that a finite-dimensional complex Lie algebra is semisimple exactly when its Killing form is nondegenerate. To see why nondegeneracy implies semisimplicity, any nonzero solvable ideal has a last nonzero derived term which is an Abelian ideal of $L$. The previous paragraph would put it in the radical of the Killing form, a contradiction.

If $L$ is a semisimple ideal of $M$, the restriction ${\kappa }_M{|}_L={\kappa }_L$ is nondegenerate. Put $J=L^{{\perp }_{{\kappa }_M}}$. This is an ideal of $M$ by the argument above, and nondegeneracy on $L$ gives $L\cap J=0$ and $M=L+J$. One way to check the latter is to restrict the pairing map $M\to L^{\ast }$: its restriction to $L$ is an isomorphism, so every vector of $M$ can be adjusted by an element of $L$ to lie in its kernel $J$. Since both are ideals, $[L,J]\subseteq L\cap J=0$. HenceApply this to the supplied extension by the derivation. If $\delta =0$, it is $\mathrm{ad}0$. Otherwise $D$ and $J$ are one-dimensional, and projection $J\to D$ is an isomorphism. Choose its vector with $D$-component $\delta$, writing it as $j=\delta -x$ for $x\in L$. Because $[J,L]=0$, for every $l\in L$ one has $0=[j,l]=\delta (l)-[x,l]$. ThusThis also handles the sign in the extension bracket explicitly.