For fixed $X\in L(G)$, set $b(t)=\varphi (e^{tX})$. The homomorphism property gives $b(s+t)=b(s)b(t)$ and $b(0)=I$. Its initial derivative is $L(\varphi )(X)=D{\varphi }_I(X)$. The one-parameter subgroup uniqueness theorem givesFor the Adjoint representation, ${\mathrm{Ad}}_{g}Y=gY{g}^{-1}$. ThusThe differential $L(\varphi )$ is a real linear map of Lie algebras; the bracket identity follows by differentiating the compatibility of conjugation with $\varphi$.

For a Matrix Lie group, $T_{g}G$ consists of derivatives ${\gamma }^{\prime }(0)$ of differentiable curves in $G$ with $\gamma (0)=g$, viewed inside $M_n(\mathbb{C})$ as a real vector space. At the identity, multiplying two such curves adds their derivatives, and reparametrizing a curve multiplies its derivative by any real scalar. Thus $L(G)=T_{I}G$ is a real vector subspace.

If $Y\in L(G)$ and $g\in G$, conjugating a curve with derivative $Y$ proves $gY{g}^{-1}\in L(G)$. Let $g(s)$ have $g(0)=I$ and $g^{\prime }(0)=X\in L(G)$. Differentiating the curve $g(s)Yg(s{)}^{-1}$ inside the finite-dimensional, hence closed, vector subspace $L(G)$ givesThe matrix commutator is real bilinear, antisymmetric, and satisfies the Jacobi identity by associativity of matrix multiplication. This proves that $L(G)$ is a real Lie algebra.

The exponential-map theorem for a linear Lie group says $\exp X\in G$ whenever $X\in L(G)$. Applying it to every real multiple of $X$ gives $\exp (tX)\in G$ for every real $t$. Conversely that curve has derivative $X$ at zero, so $X\in L(G)$. Hence

The full-cycle formula says that precisely the nonhook shapes have zero value on a 6-cycle. These are $(4,2),(3,3),(3,2,1),(2,2,1,1),(2,2,2)$.

Use the standard two-row permutation-character rule ${\pi }^{(6-r,r)}={\chi }^{(6)}+{\chi }^{(5,1)}+\cdots +{\chi }^{(6-r,r)}$. A transposition has four fixed points, so ${\chi }^{(5,1)}=3$. It fixes $\left(\genfrac{}{}{0ex}{}{4}{2}\right)+1=7$ two-subsets, giving ${\chi }^{(4,2)}=7-1-3=3$. It fixes $\left(\genfrac{}{}{0ex}{}{4}{3}\right)+\left(\genfrac{}{}{0ex}{}{4}{1}\right)=8$ three-subsets, giving ${\chi }^{(3,3)}=8-1-3-3=1$.

Conjugating a partition tensors its character with the sign character. Hence the values on a transposition for the conjugate shapes are $-3$ and $-1$. The shape $(3,2,1)$ is self-conjugate, so its character equals its own sign twist and is zero on every odd permutation. The complete candidate table is therefore:

Thus the unique answer is

Irreducible character orthogonality gives $⟨\chi ,\chi ⟩=1$. Under the proposed vanishing condition, only the identity contributes, soBut $\chi (1)$ is an integer dimension and $720$ is not a square. Thus no such partition exists.

There is more than one solution: the trivial representation $\lambda =(6)$ has character $1$ everywhere and the sign representation $\lambda =(1^6)$ has character $\pm 1$ everywhere. ThusIn fact they are the only ones. The full-cycle formula in the preceding discussion forces any candidate to be a hook $(6-r,1^r)$. For $r=1$, the standard character is the number of fixed points minus one, so it vanishes on cycle type $(5,1)$; its sign twist $r=4$ does too. For $r=2$, the hook representation is the exterior square of the standard representation and has character $\frac{1}{2}(\chi (g{)}^2-\chi (g^2))$. At cycle type $(4,1,1)$ both standard-character values are one, so this character is zero. Its sign twist $r=3$ also vanishes there. Only $r=0,5$ remain.

Writing matrices in $2\times 2$ blocks, the defining relation givesThe trace is automatically zero. The relation is linear, and if $as=-s{a}^T$, $bs=-s{b}^T$, thenThus $L$ is closed under the commutator and is a Lie subalgebra of ${\mathfrak{sl}}_4(F)$.

Its diagonal subalgebra isPut ${\epsilon }_1(D(a,b))=a$, ${\epsilon }_2(D(a,b))=b$. Since $[D,E_{ij}]=(D_{ii}-D_{jj})E_{ij}$, the required block decomposition isThe four lines have respective characters ${\epsilon }_1-{\epsilon }_2$, $-{\epsilon }_1+{\epsilon }_2$, ${\epsilon }_1+{\epsilon }_2$, $-{\epsilon }_1-{\epsilon }_2$. They are distinct and nonzero because $\mathrm{char}F\ne 2$. The two-dimensional summand $H$ has two trivial composition factors. These five submodules therefore have exactly the factor-grouping properties of part (b). Their dimensions sum to $6$, the dimension obtained from four free entries in $A$ and one each in the skew-symmetric blocks $B,C$.

The flag makes every composition factor one-dimensional, described by a character $\lambda :H\to F$. The Jordan–Hölder theorem for modules for finite-length modules says that the multiset of simple composition factors is independent of the chosen composition series.

Here is a construction valid over every $F$. For $h\in H$, every generalized eigenspace of $\rho (h)$ is an $H$-submodule. Indeed $\mathrm{ad}(h)$ is nilpotent on $H$, say of exponent $c$, andIf $v$ is killed by $(\rho (h)-aI{)}^{N_0}$, take $N\ge N_0+c$. Terms with $j\ge c$ vanish by adjoint nilpotence, and all other terms kill $v$. This proves invariance of the generalized eigenspace.

The flag also makes the characteristic polynomial of every $\rho (h)$ split over $F$. Whenever two composition characters differ, choose an $h$ on which they differ and split the module into the distinct generalized eigenspaces of $\rho (h)$. Repeat within each summand until all its composition factors have the same character. Every nontrivial split lowers dimensions, so the process terminates. Finally combine summands carrying the same character. Jordan–Hölder ensures that the resulting distinct blocks have distinct factor types, givingThis recursive method does not assume that a single element separates every character, which could fail over a finite field.

Choose a basis $(u,v)$ with $U=Fu$. The representation matrices areThe one-dimensional modules $U,V/U$ have characters $\lambda ,\mu$. If they coincide, these modules are isomorphic and the first alternative holds.

Otherwise choose $h$ with $\delta (h)\ne 0$, where $\delta =\lambda -\mu$. Replacing $v$ by $v-a(h)u/\delta (h)$ makes the upper-right entry of $\rho (h)$ zero. The representation identity on commutators now gives, for every $x\in H$,Iteration yields $a((\mathrm{ad}h{)}^{r}x)=\delta (h{)}^{r}a(x)$. Nilpotence of $H$ implies $(\mathrm{ad}h{)}^{r}x=0$ for sufficiently large $r$, by its lower central series. Since $\delta (h)\ne 0$, it follows that $a(x)=0$ for all $x$. The new line $W=Fv$ is therefore a submodule, andThe argument works over the stated arbitrary field, including positive characteristic.

Let $\pi :L\to L/I$ be the quotient map. A full flag of ideals in $L/I$ lifts by inverse images to a flag from $I$ to $L$, with dimensions increasing by one. Prepending $0<I$ supplies the required full flag in $L$. Thus supersolubility lifts across a one-dimensional ideal.

Over $\mathbb{C}$, a nonzero soluble $L$ has a one-dimensional ideal: apply part (a) to its adjoint image acting on $L$. A common eigenvector spans an adjoint-invariant line, which is an ideal. The quotient is soluble and has smaller dimension, so induction and the lifting result give a full flag of ideals. Therefore a soluble complex Lie algebra is supersoluble.

Conversely this implication needs no field restriction. If $L$ has a full flag, its first nonzero ideal $I$ is one-dimensional and abelian, and $L/I$ has a full flag. By induction $L/I$ is soluble, so some term of the derived series of $L$ lies in $I$; its next derived term is zero. Hence

We prove Lie's theorem in the required form by induction on $\dim L$. If $L=0$, any nonzero vector works. Otherwise solvability gives $[L,L]\ne L$, so choose a codimension-one ideal $K\supseteq [L,L]$ and $x\in L\setminus K$. By induction there are $v\ne 0$ and a linear functional $\lambda :K\to \mathbb{C}$ with $av=\lambda (a)v$ for all $a\in K$.

Let $W=\mathrm{span}\{v,xv,x^{2}v,\dots \}$. Choose the cyclic basis $v,xv,\dots ,x^{r-1}v$ up to its first linear dependence. The commutator identity $a{x}^j=xa{x}^{j-1}+[a,x]x^{j-1}$, with $[a,x]\in K$, proves inductively thatHence $W$ is invariant under $K$ and $x$, and every $a\in K$ has trace $r\lambda (a)$ on $W$. In particular ${\mathrm{tr}}_W([x,a])=0$ implies $r\lambda ([x,a])=0$. Characteristic zero gives $\lambda ([x,a])=0$.

The simultaneous weight space $V_{\lambda }=\{w:aw=\lambda (a)w\forall a\in K\}$ is nonzero and stable under $x$, becauseOver $\mathbb{C}$, the restriction of $x$ to this nonzero finite-dimensional space has an eigenvector. That vector is an eigenvector of both $x$ and every $a\in K$, hence of every element of $L$.

The four constructed submodules have the distinct shapes $(n-3,3)$, $(n-2,2)$, $(n-1,1)$ and $(n)$, so they are pairwise nonisomorphic irreducible characteristic-zero modules. Complete reducibility therefore makes their sum direct. The given decomposition of $M^{(n-2,2)}$ and the dimension identity in (a) show their dimensions sum to $\dim M^{(n-3,3)}$. Their direct sum is consequently the whole module:

For the standard tableau with second row $2,4$, the first entries in the top row are $1,3$, so its nontrivial columns are $(1,2)$ and $(3,4)$. HenceIn ${\varphi }_2(e_t)$ the coefficient of $x_{24n}$ is one: among those four pairs only $\{2,4\}$ is contained in $\{2,4,n\}$, since $n\ge 6$. Thus the image is nonzero. Its restriction to the characteristic-zero simple module $S^{(n-2,2)}$ has zero kernel, giving the requested isomorphic submodule.

For $S^{(n-1,1)}$, choose the standard tableau with lower entry two and top-left entry one. Its polytabloid is $v_2-v_1$. Each three-subset containing $a$ occurs twice in ${\varphi }_2{\varphi }_1(v_a)$, once for each choice of its other two entries as the intermediate $b$. Consequently the coefficient of $x_{23n}$ in ${\varphi }_2{\varphi }_1(v_2-v_1)$ is two. This is nonzero in characteristic zero. Simplicity again makes the restriction injective, giving a copy of $S^{(n-1,1)}$.

A permutation sends $v_a$ to $v_{ga}$, $w_{ab}$ to $w_{ga,gb}$ and $x_{abc}$ to $x_{ga,gb,gc}$. Relabeling the summation indices givesLinearity therefore proves both maps are module homomorphisms. The map denoted $\varphi$ in the second displayed source formula is ${\varphi }_2$.

For a tableau $t$, let $C_t$ be its column stabilizer and $e_t={\sum }_{\sigma \in C_t}\mathrm{sgn}(\sigma )\{\sigma t\}$ its polytabloid. The Specht module is the span of all such polytabloids inside the tabloid permutation module.

A tabloid with second row of size $k$ is determined by that subset, so $\dim M^{(n-k,k)}=\left(\genfrac{}{}{0ex}{}{n}{k}\right)$. The two-row hook formula givesAt $k=3$ this provesBy Maschke's theorem, characteristic-zero permutation modules are completely reducible, so their Specht submodules are summands. The nonzero sum of all three-subset tabloids is fixed by every permutation; its span is a copy of the trivial module $S^{(n)}$.