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Every $2\times 2$ matrix has the unique decompositionBoth summands define invariant subspaces for the Adjoint representation of a Lie algebra. Scalar matrices commute with everything, so $\mathbb{C}I$ is $L_0$. The traceless matrices form ${\mathfrak{sl}}_2$ with basis $e,h,f$ of weights $2,0,-2$. Their brackets satisfy $[e,f]=h$, $[h,e]=2e$, $[h,f]=-2f$; the raising and lowering operations connect all three weight lines. This is the simple module $L_2$. ThereforeAs a check, its zero-weight space has dimension two and its one-weight space is zero, agreeing with part (ii).

In the simple module $L_m$, weight $0$ occurs once if $m$ is even and never if $m$ is odd. Weight $1$ occurs once if $m$ is odd and never if $m$ is even. Thus each simple summand contributes exactly one to $\dim V_0+\dim V_1$.

Complete reducibility of semisimple Lie algebra representations writes $V={⨁}_m{L}_m^{\oplus a_m}$. Taking the two weight spaces commutes with this finite direct sum, soThe right side counts irreducible summands with multiplicity, independently of a chosen decomposition.

By the classification of finite-dimensional sl2 representations over $\mathbb{C}$, every finite-dimensional module is a direct sum of simple modules $L_m$, indexed by integers $m\ge 0$. The module $L_m$ has dimension $m+1$ and a basis $v_0,\dots ,v_m$ withwith out-of-range vectors interpreted as zero. In particular $v_0$ is a highest weight vector of weight $m$. The weights are $m,m-2,\dots ,-m$, each with multiplicity one. The multiplicities of the $L_m$ in a finite module determine that module up to isomorphism.

Suppose $\mathrm{ad}(x)$ has a nonzero eigenvalue $\lambda$ and corresponding vector $y$. Then $[x,y]=\lambda y$, soFor a diagonalizable endomorphism $T$, its eigenspace decomposition shows $\ker T^2=\ker T$. The displayed equations therefore contradict diagonalizability of $\mathrm{ad}(y)$.

Thus no $\mathrm{ad}(x)$ has a nonzero eigenvalue. Each is diagonalizable over the algebraically closed field, so each is the zero endomorphism. Hence $[x,z]=0$ for every $x,z\in \mathfrak{g}$, proving

If $\mathfrak{g}$ is nilpotent, every Lie subalgebra $\mathfrak{a}$ is nilpotent, since ${\mathfrak{a}}^j\subseteq {\mathfrak{g}}^j$ in the respective lower central series of a Lie algebra. A two-dimensional nilpotent algebra has nonzero centre by part (i). If $0\ne z$ is central, complete it to a basis $(z,u)$; every bracket of these basis elements is zero. Thus $\mathfrak{a}$ is abelian.

Conversely, suppose $\mathfrak{g}$ is not nilpotent. By Engel's theorem, some $\mathrm{ad}(x)$ is not nilpotent. Since the field is algebraically closed, it has an eigenvalue $\lambda \ne 0$ with eigenvector $y\ne 0$. ThusThe vectors $x,y$ are independent: otherwise their bracket would be zero. Their span is therefore a two-dimensional Lie subalgebra and it is nonabelian. This contradicts the hypothesis. Hence

Fix $x$ with $x^N=0$ as a matrix. On the full matrix algebra define $L_x(y)=xy$ and $R_x(y)=yx$. These endomorphisms commute and each has $N$th power zero. The Adjoint representation of a Lie algebra is $\mathrm{ad}(x)=L_x-R_x$. The binomial theorem givesFor $m=2N-1$, each summand has either $j\ge N$ or $m-j\ge N$, and hence vanishes. Therefore $\mathrm{ad}(x)$ is nilpotent on the full matrix algebra and on its invariant subspace $\mathfrak{g}$.

Engel's theorem, in its adjoint form, states that a finite-dimensional Lie algebra is nilpotent if every $\mathrm{ad}(x)$ is a nilpotent endomorphism. Applying it to all $x\in \mathfrak{g}$ provesThis is distinct from assuming that an arbitrary representation of a nilpotent algebra must act by nilpotent matrices, which is false.

Write the lower central series of a Lie algebra as ${\mathfrak{g}}^1=\mathfrak{g}$ and ${\mathfrak{g}}^{j+1}=[\mathfrak{g},{\mathfrak{g}}^j]$. Since the nilpotent Lie algebra is nonzero, there is a last nonzero term ${\mathfrak{g}}^c$, with ${\mathfrak{g}}^{c+1}=0$. Thus $[\mathfrak{g},{\mathfrak{g}}^c]=0$, soConsequently the centre is nonzero. This argument works over every field and includes an abelian algebra, for which $c=1$.

A tabloid is the row-equivalence class $\{t\}$ of a tableau filled bijectively with $1,\dots ,n$: order inside each row is forgotten, but the rows remain distinguished. The permutation module $M^{\lambda }$ has these tabloids as an $F$-basis, with ${\Sigma }_n$ permuting their entries. Let $C_t$ be the column stabilizer of $t$ and defineThe Specht module $S^{\lambda }$ is the submodule spanned by all polytabloids $e_t$.

If a module homomorphism $\varphi :M^{\lambda }\to M^{\mu }$ is nonzero on $S^{\lambda }$, it is nonzero on some spanning $e_t$. Equivariance gives $0\ne \varphi (e_t)={\kappa }_t\varphi (\{t\})$, so ${\kappa }_t{M}^{\mu }\ne 0$. The allowed criterion yields $\lambda \ge \mu$.

In characteristic zero, Maschke's theorem gives an equivariant projection $M^{\lambda }\to S^{\lambda }$. If $S^{\lambda }\cong S^{\mu }$, compose that projection with the isomorphism and the inclusion into $M^{\mu }$. The preceding argument gives $\lambda \ge \mu$. Reversing the roles gives $\mu \ge \lambda$, so antisymmetry of dominance givesFor a map defined only on a Specht module in arbitrary characteristic, dominance need not hold. Take $F={\mathbb{F}}_2$, $n=2$, $\lambda =(1,1)$ and $\mu =(2)$. The first Specht module is the one-dimensional span ofThe transposition fixes $e$, so this is the trivial module. The one-row permutation module $M^{(2)}$ is also trivial and one-dimensional. Sending $e$ to its single tabloid is a nonzero equivariant map, yet $(1,1)\ngeqq (2)$. In this characteristic such a map need not extend to the surrounding permutation module; that is exactly where the earlier proof ceases to apply.

For a Young diagram cell $(i,j)$, its hook length is one plus the number of cells to its right and below. The hook-length formula isApplying it to the two shapes givesThe shapes here require $n\ge 4$.

A hook tabloid remembers its first-row set and the ordered singleton entries below it. The proposed map keeps that first-row set and merges the remaining entries into one unordered row. Thus it is well-defined on tabloids. Permuting entries before merging gives the same tabloid as merging before permuting, so it is a module homomorphism. Every target tabloid is reached by putting its second-row entries, in any order, into the singleton rows. Therefore it is surjective.

For $r=2$, the strictly dominating partitions of $\lambda =(n-2,1,1)$ are exactly $(n),(n-1,1),(n-2,2)$. Indeed dominance forces the first part to be at least $n-2$, and the possibilities for the remaining sum give precisely those shapes or $\lambda$ itself.

In characteristic zero, Maschke's theorem makes the permutation module semisimple, and Young's rule gives one copy of $S^{\lambda }$ and only strictly dominating additional shapes. The trivial multiplicity is one, since a transitive permutation module has a one-dimensional invariant space. For the standard representation $S^{(n-1,1)}$, Frobenius reciprocity identifies its multiplicity with the dimension of fixed vectors under the row stabilizer $S_{n-2}$. In its realization $\{(z_1,\dots ,z_n):\sum z_i=0\}$, those fixed vectors have the form $(c,\dots ,c,a,b)$ with $(n-2)c+a+b=0$, a two-dimensional space. Thus its multiplicity is two.

Finally $\dim M^{\lambda }=n(n-1)$. Subtracting the dimensions of one $S^{\lambda }$, one trivial module and two standard modules leaves $n(n-3)/2$, exactly one $S^{(n-2,2)}$. ThereforeThe additional multiplicities are $1,2,1$ respectively. If the printed weak summation index is read literally alongside its separate $S^{\lambda }$, then $m_{\lambda ,\lambda }=0$; the total multiplicity of $S^{\lambda }$ is one.

A reduced crystallographic root system $\Phi$ is a finite spanning subset of a real Euclidean vector space, excluding zero, such that $\Phi \cap \mathbb{R}\alpha =\{\pm \alpha \}$, each reflection $s_{\alpha }(v)=v-2(v,\alpha )\alpha /(\alpha ,\alpha )$ preserves $\Phi$, and all Cartan integers $⟨\beta ,\alpha ⟩=2(\beta ,\alpha )/(\alpha ,\alpha )$ are integers. A base is a vector-space basis $\Delta \subset \Phi$ in which each root has integer coordinates either all nonnegative or all nonpositive. The Weyl group is generated by the root reflections.

For nonparallel roots put $p=⟨\beta ,\alpha ⟩$ and $q=⟨\alpha ,\beta ⟩$. Then $pq=4{\cos }^2\theta$ is an integer strictly below $4$ and at least zero. ThereforeIf $(\alpha ,\beta )>0$, the positive integers $p,q$ have product $1,2$ or $3$, so at least one is one. If $p=1$, reflection gives $\beta -\alpha \in \Phi$ and hence its negative. If $q=1$, reflection directly gives $\alpha -\beta \in \Phi$. This proves the positive-inner-product assertion.

To prove that an $\alpha$-string has no gaps, suppose consecutive occurring integer positions $a<b$ in $\beta +t\alpha$ had $b-a\ge 2$. If $(\beta +a\alpha ,\alpha )<0$, apply the previous result to $\beta +a\alpha$ and $-\alpha$ to obtain the missing root at $a+1$. Thus this inner product must be nonnegative. Similarly the absence of a root at $b-1$ forces $(\beta +b\alpha ,\alpha )\le 0$. But the latter inner product exceeds the former by $(b-a)(\alpha ,\alpha )>0$, impossible. Thus every intermediate integer position occurs.

Write the complete finite string as $\beta -r\alpha ,\dots ,\beta +s\alpha$. Reflection maps position $t$ to $-p-t$, so it interchanges the two endpoints. HenceHere $r,s$ are the maximal endpoint distances, not arbitrary shorter distances mentioned in the existence assertion.

Use the Cartan-matrix convention $C_{ij}=⟨{\alpha }_i,{\alpha }_j⟩$. The displayed matrix is realized bywith the standard orthonormal coordinates. Closure under the two simple reflections givesThe positive roots are ${\alpha }_1,{\alpha }_2,{\alpha }_1+{\alpha }_2,2{\alpha }_1+{\alpha }_2$. Conversely the standard fact that every root is Weyl-conjugate to a simple root, or successive height reduction by simple reflections, shows these two reflection orbits exhaust the roots. This is the rank-two system $B_2$, also called $C_2$ after exchanging the length convention. With a transposed Cartan-matrix convention, interchange which simple-root label is short; the geometric system and Weyl group are unchanged.

The reflection in $e_2$ flips the second coordinate, while the reflection in $e_1-e_2$ swaps the two coordinates. These generate all eight signed permutation matrices in two coordinates. Thuswhere $D_8$ means the symmetry group of a square.

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.