Solvable and Nilpotent Lie Algebras

For every subspace $A\subseteq \mathfrak{g}$, linearity and bracket preservation give $f([A,A])=[f(A),f(A)]$: each spanning bracket maps to a spanning bracket, and every bracket on the right has a preimage because $f$ is surjective.

Starting with $f(\mathfrak{g})=\mathfrak{g}$, suppose $f({\mathfrak{g}}^{(j)})={\mathfrak{g}}^{(j)}$. Then [L1] and step 1.1 give $f({\mathfrak{g}}^{(j+1)})=[{\mathfrak{g}}^{(j)},{\mathfrak{g}}^{(j)}]={\mathfrak{g}}^{(j+1)}$. Finite induction proves the equality at the given index $r$; [L1] already supplies ideality.

For subspaces $A,B\subseteq \mathfrak{g}$, bracket preservation, linearity, and surjectivity give $f([A,B])=[f(A),f(B)]$.

The equality $f({\gamma }_1)=\mathfrak{g}={\gamma }_1$ is immediate. If $f({\gamma }_j)={\gamma }_j$, then [L1] and step 1.1 give $f({\gamma }_{j+1})=f([\mathfrak{g},{\gamma }_j])=[\mathfrak{g},{\gamma }_j]={\gamma }_{j+1}$. Finite induction reaches the given $r$, and [L1] supplies ideality.

Assume ${\gamma }_{c+1}=0$. We prove ${\gamma }_{c+1-r}\subseteq Z_r$ for $0\le r\le c$. At $r=0$ this is the assumed containment in $Z_0=0$. If it holds at $r$, then $x\in {\gamma }_{c-r}$ implies $[\mathfrak{g},x]\subseteq {\gamma }_{c+1-r}\subseteq Z_r$, so [L2] gives $x\in Z_{r+1}$. At $r=c$ we obtain $\mathfrak{g}={\gamma }_1\subseteq Z_c$, hence equality.

Conversely assume $Z_c=\mathfrak{g}$. Starting from ${\gamma }_1=\mathfrak{g}=Z_c$, induction gives ${\gamma }_{r+1}\subseteq Z_{c-r}$ for $0\le r\le c$: if ${\gamma }_{r+1}\subseteq Z_{c-r}$, then [L1] and [L2] give ${\gamma }_{r+2}=[\mathfrak{g},{\gamma }_{r+1}]\subseteq Z_{c-r-1}$. At $r=c$ this yields ${\gamma }_{c+1}\subseteq Z_0=0$, so $\mathfrak{g}$ is nilpotent. The argument also covers $c=0$, when both conditions say $\mathfrak{g}=0$.

We first prove $[{\gamma }_p,{\gamma }_q]\subseteq {\gamma }_{p+q}$ for all $p,q\ge 1$. For $p=1$ this is [L2]. If it holds for $p-1$ and every second index, Jacobi gives $[[\mathfrak{g},{\gamma }_{p-1}],{\gamma }_q]\subseteq [\mathfrak{g},[{\gamma }_{p-1},{\gamma }_q]]+[{\gamma }_{p-1},[\mathfrak{g},{\gamma }_q]]\subseteq [\mathfrak{g},{\gamma }_{p+q-1}]+[{\gamma }_{p-1},{\gamma }_{q+1}]\subseteq {\gamma }_{p+q}$.

At $r=0$, ${\mathfrak{g}}^{(0)}=\mathfrak{g}={\gamma }_1$. If ${\mathfrak{g}}^{(r)}\subseteq {\gamma }_{2^r}$, then [L1] and step 1.1 imply ${\mathfrak{g}}^{(r+1)}\subseteq [{\gamma }_{2^r},{\gamma }_{2^r}]\subseteq {\gamma }_{2^{r+1}}$. Thus the displayed containment holds for every $r$.

If $\mathfrak{g}$ is nilpotent, choose a given bound $c$ with ${\gamma }_{c+1}=0$ as in [L2]. Taking $r=c+1$ gives $2^r\ge c+1$, so descent of the lower series and step 2.1 yield ${\mathfrak{g}}^{(r)}\subseteq {\gamma }_{2^r}\subseteq {\gamma }_{c+1}=0$. Hence $\mathfrak{g}$ is solvable by [L1]. This includes $\mathfrak{g}=0$ and uses only the supplied finite bound.

Induction on $r$ gives ${\mathfrak{h}}^{(r)}\subseteq {\mathfrak{g}}^{(r)}$: it is clear at $r=0$, and bracketing a subspace with itself preserves containment. Therefore a vanishing derived term of $\mathfrak{g}$ forces the corresponding term of $\mathfrak{h}$ to vanish.

Bracket preservation and surjectivity of the canonical map in [L2] give $(\mathfrak{g}\mathrm{/}\mathfrak{i}{)}^{(r)}=\pi ({\mathfrak{g}}^{(r)})$ for every $r$. Thus every solvable quotient of $\mathfrak{g}$ terminates no later than $\mathfrak{g}$; [L3] records the same calculation for any homomorphic image.

Suppose $m=\mathrm{dl}(\mathfrak{g}\mathrm{/}\mathfrak{i})$ and $n=\mathrm{dl}(\mathfrak{i})$. Step 1.2 gives $\pi ({\mathfrak{g}}^{(m)})=0$, so ${\mathfrak{g}}^{(m)}\subseteq \ker \pi =\mathfrak{i}$. Iterating the derived operation $n$ further times gives ${\mathfrak{g}}^{(m+n)}=({\mathfrak{g}}^{(m)}{)}^{(n)}\subseteq {\mathfrak{i}}^{(n)}=0$. This proves solvability and the bound, including $m=0$ or $n=0$.

Induction gives ${\gamma }_r(\mathfrak{h})\subseteq {\gamma }_r(\mathfrak{g})$ for all $r\ge 1$, because $[\mathfrak{h},A]\subseteq [\mathfrak{g},A]$. Thus any lower-series term vanishing in $\mathfrak{g}$ also vanishes in $\mathfrak{h}$.

If $\pi :\mathfrak{g}\to \mathfrak{g}\mathrm{/}\mathfrak{i}$ is the map in [L3], surjectivity and bracket preservation give ${\gamma }_r(\mathfrak{g}\mathrm{/}\mathfrak{i})=\pi ({\gamma }_r(\mathfrak{g}))$ by induction. Hence quotients inherit lower-series termination.

Componentwise bracketing in [L4] gives ${\gamma }_r({\prod }_{j\in J}{\mathfrak{g}}_j)={\prod }_{j\in J}{\gamma }_r({\mathfrak{g}}_j)$ for every $r\ge 1$. If $J$ is nonempty, let $c={\max }_{j\in J}\mathrm{cl}({\mathfrak{g}}_j)$. For $c=0$ every factor, and hence the product, is zero. For $c\ge 1$, the product's $(c+1)$st term is zero, while its $c$th term is nonzero in a factor attaining the maximum. Thus [L2] gives class exactly $c$ in either case. If $J$ is empty, [L4] identifies the product with zero and [L2] gives class zero.

Surjectivity and bracket preservation in [L2] give $\pi ({\gamma }_r(\mathfrak{g}))={\gamma }_r(\mathfrak{g}\mathrm{/}\mathfrak{i})$ by induction on $r$. Since the quotient has class $c$, its $(c+1)$st term vanishes, and therefore ${\gamma }_{c+1}(\mathfrak{g})\subseteq \ker \pi =\mathfrak{i}$.

Now [L1], step 1.1, and centrality [L3] give ${\gamma }_{c+2}(\mathfrak{g})=[\mathfrak{g},{\gamma }_{c+1}(\mathfrak{g})]\subseteq [\mathfrak{g},\mathfrak{i}]=0$. Thus $\mathfrak{g}$ has class at most $c+1$. When the quotient is zero ($c=0$), this says precisely that the central algebra $\mathfrak{g}=\mathfrak{i}$ is abelian.

Because ${\gamma }_1(\mathfrak{g})=\mathfrak{g}\mathrm{\ne }0$ and the series terminates by [L1], there is a largest index $c\ge 1$ with ${\gamma }_c(\mathfrak{g})\mathrm{\ne }0$. Its successor satisfies $[\mathfrak{g},{\gamma }_c]={\gamma }_{c+1}=0$, so [L2] gives $0\mathrm{\ne }{\gamma }_c\subseteq Z(\mathfrak{g})$.

Hence the center contains the displayed nonzero subspace and is itself nonzero. The hypothesis $\mathfrak{g}\mathrm{\ne }0$ is essential: the zero algebra has zero center. No choice of a basis or of a central element is needed.

If $\mathfrak{g}=0$, every nonzero $v\in V$ is annihilated by $\mathfrak{g}$, so the assertion holds.

Assume $\mathfrak{g}\mathrm{\ne }0$ and that the assertion holds for every nil Lie algebra of operators of dimension strictly smaller than $\dim \mathfrak{g}$, acting on any nonzero finite-dimensional module.

Choose a proper subalgebra $\mathfrak{h}<\mathfrak{g}$ of maximal dimension; this exists because $0$ is proper and the possible dimensions form a nonempty finite set. For $H\in \mathfrak{h}$, take $m>0$ with $H^m=0$ by [L1]. On $\mathrm{End}(V)$, ${\mathrm{ad}}_H=L_H-R_H$, where $L_H$ and $R_H$ commute, and every term of $(L_H-R_H{)}^{2m-1}$ contains either $L_H^m$ or $R_H^m$; hence ${\mathrm{ad}}_H$ is nilpotent. Its restrictions and induced quotient operators are nilpotent as well.

The adjoint action of $\mathfrak{h}$ preserves $\mathfrak{h}$, so [L2] gives an action on the nonzero space $\mathfrak{g}\mathrm{/}\mathfrak{h}$. By step 1.3 it is nil, and step 1.2 supplies a nonzero coset $x+\mathfrak{h}$ killed by $\mathfrak{h}$. Thus $x\mathrm{\notin }\mathfrak{h}$ and $[\mathfrak{h},x]\subseteq \mathfrak{h}$, so the normalizer of $\mathfrak{h}$ strictly contains $\mathfrak{h}$.

The normalizer is a subalgebra; maximality of $\mathfrak{h}$ in step 1.3 and step 2.1 therefore make it all of $\mathfrak{g}$, so $\mathfrak{h}$ is an ideal. Moreover $\mathfrak{g}\mathrm{/}\mathfrak{h}$ is one-dimensional: otherwise the inverse image of the one-dimensional subalgebra spanned by any nonzero quotient vector would be strictly between $\mathfrak{h}$ and $\mathfrak{g}$. Hence $\mathfrak{g}=\mathfrak{h}\oplus kx$.

Apply step 1.2 to $\mathfrak{h}$ acting on $V$. Its common kernel $V_0=\{v\in V:\mathfrak{h}v=0\}$ is nonzero. It is $x$-stable, because for $H\in \mathfrak{h}$ and $v\in V_0$, $H(xv)=x(Hv)+[H,x]v=0$ by ideality from step 3.1.

The restriction of $x$ to nonzero finite-dimensional $V_0$ is nilpotent by [L1]. Its kernel is nonzero: if it were zero, rank-nullity [L3] would make $x{\mathrm{\mid }}_{V_0}$ injective, hence every positive power injective, contradicting nilpotence on $V_0\mathrm{\ne }0$. Choose $0\mathrm{\ne }v\in \ker (x{\mathrm{\mid }}_{V_0})$. Then $\mathfrak{h}v=0$, $xv=0$, and step 3.1 gives $\mathfrak{g}v=0$. This is a single finite existential choice, not an application of Choice.

If $V=0$, the empty flag has the required property and the empty basis gives the matrix assertion.

Assume $V\mathrm{\ne }0$ and the theorem for all nil modules of dimension smaller than $\dim V$.

By [L1], choose $0\mathrm{\ne }v\in V$ with $\mathfrak{g}v=0$, and set $V_1=kv$. This is an invariant one-dimensional subspace annihilated by $\mathfrak{g}$.

Every induced operator on $V\mathrm{/}{V}_1$ is nilpotent, since a power of $\rho (x)$ that vanishes on $V$ also vanishes on the quotient. By [L2] and step 1.2, the quotient has a flag $0={\bar{V}}_1\subset \cdots \subset {\bar{V}}_n=V\mathrm{/}{V}_1$ with $\mathfrak{g}{\bar{V}}_i\subseteq {\bar{V}}_{i-1}$. Taking inverse images under the projection [L3] and adjoining $0\subset V_1$ gives the required flag of $V$.

Choose vectors adapted to the flag, in the finite sequential sense. The containment $\rho (\mathfrak{g})V_i\subseteq V_{i-1}$ says every matrix sends the $i$th basis vector into the span of earlier vectors, hence is strictly upper triangular in this ordering. Conversely a common strictly upper-triangular basis supplies exactly this flag. The zero case was covered in step 1.1.

Suppose ${\gamma }_{c+1}(\mathfrak{g})=0$. For $x,y\in \mathfrak{g}$, the vector $({\mathrm{ad}}_x{)}^c(y)$ is a left-nested bracket with $c$ copies of $x$, hence lies in ${\gamma }_{c+1}=0$ by [L1] and [L3]. Thus every ${\mathrm{ad}}_x$ is nilpotent. This also covers $\mathfrak{g}=0$.

Conversely suppose every ${\mathrm{ad}}_x$ is nilpotent. The adjoint representation in [L3] is nil, so [L2] gives $0=V_0\subset V_1\subset \cdots \subset V_n=\mathfrak{g}$ with $[\mathfrak{g},V_i]\subseteq V_{i-1}$. Induction then gives ${\gamma }_{r+1}(\mathfrak{g})\subseteq V_{n-r}$ for $0\le r\le n$, and in particular ${\gamma }_{n+1}\subseteq V_0=0$. Hence $\mathfrak{g}$ is nilpotent by [L1].

Apply [L1] to the adjoint representation. If $0=V_0\subset \cdots \subset V_m=\mathfrak{g}$ is its lowered flag, put $A_i=V_{m-i}$. Then $A_0=\mathfrak{g}$, $A_m=0$, and $[\mathfrak{g},A_i]\subseteq A_{i+1}$. In particular [L2] also confirms that $\mathfrak{g}$ is nilpotent.

We prove $A_{m-r}\subseteq Z_r$ for $0\le r\le m$. At $r=0$ this is $A_m=0=Z_0$. If it holds at $r$, then $[\mathfrak{g},A_{m-r-1}]\subseteq A_{m-r}\subseteq Z_r$, so [L3] gives $A_{m-r-1}\subseteq Z_{r+1}$. At $r=m$ this yields $\mathfrak{g}=A_0\subseteq Z_m$, hence equality. For $\mathfrak{g}=0$, take $m=0$.

If $\dim \mathfrak{g}=1$, the only proper subalgebra is $0$, which is itself an ideal of codimension one. The case $\mathfrak{g}=0$ has no proper subalgebra and is vacuous.

Assume $\dim \mathfrak{g}>1$ and the assertion for all nilpotent Lie algebras of smaller dimension.

By [L1], choose $0\mathrm{\ne }z\in Z(\mathfrak{g})$ and put $\mathfrak{a}=kz$. Then $\mathfrak{a}$ is a one-dimensional central ideal.

The quotient $\bar{\mathfrak{g}}=\mathfrak{g}\mathrm{/}\mathfrak{a}$ is defined by [L3], has smaller dimension by [L4], and is nilpotent by [L2].

If $z\in \mathfrak{h}$, then $\bar{\mathfrak{h}}=\mathfrak{h}\mathrm{/}\mathfrak{a}$ is proper in $\bar{\mathfrak{g}}$. By step 1.2 there is a codimension-one ideal $\bar{\mathfrak{m}}$ containing it. Its inverse image $\mathfrak{m}$ under the quotient map is an ideal containing $\mathfrak{h}$, and [L4] gives codimension one.

If $z\mathrm{\notin }\mathfrak{h}$, set $\mathfrak{s}=\mathfrak{h}+\mathfrak{a}$; centrality makes this a subalgebra. If $\mathfrak{s}=\mathfrak{g}$, then $\mathfrak{h}$ has codimension one and $[\mathfrak{g},\mathfrak{h}]=[\mathfrak{h}+\mathfrak{a},\mathfrak{h}]\subseteq \mathfrak{h}$, so it is the required ideal. If $\mathfrak{s}$ is proper, then $\mathfrak{s}\mathrm{/}\mathfrak{a}$ is proper in $\bar{\mathfrak{g}}$; apply step 1.2 there and take the inverse image as in step 3.1.

The alternatives $z\in \mathfrak{h}$ and $z\mathrm{\notin }\mathfrak{h}$, including both subcases of the latter, exhaust all possibilities and each yields a codimension-one ideal containing $\mathfrak{h}$. The only selections were one central witness and finitely many induction witnesses, so no Choice principle is used.

The derived algebra $[\mathfrak{g},\mathfrak{g}]$ is proper. Otherwise ${\mathfrak{g}}^{(1)}=\mathfrak{g}$, so every derived term would equal the nonzero algebra $\mathfrak{g}$, contradicting solvability in [L1]. Thus the abelianization $A=\mathfrak{g}\mathrm{/}[\mathfrak{g},\mathfrak{g}]$ in [L2] is nonzero and finite-dimensional.

Choose a hyperplane $H<A$: take one nonzero vector, extend it to a finite basis, and span all basis vectors except that one. The inverse image $\mathfrak{h}$ of $H$ in $\mathfrak{g}$ contains $[\mathfrak{g},\mathfrak{g}]$, so $[\mathfrak{g},\mathfrak{h}]\subseteq [\mathfrak{g},\mathfrak{g}]\subseteq \mathfrak{h}$ and $\mathfrak{h}$ is an ideal. By [L3], its codimension equals that of $H$, namely one. Only a finite basis extension is used, so algebraic closure and characteristic zero are unnecessary and no Choice principle is invoked.

If $\mathfrak{g}=0$, any nonzero $w\in V$ is a common eigenvector with the zero functional.

Assume $\mathfrak{g}\mathrm{\ne }0$ and the theorem for solvable Lie algebras of smaller dimension.

By [L1], choose a codimension-one ideal $\mathfrak{h}$ and write $\mathfrak{g}=\mathfrak{h}\oplus kx$. Its derived series is contained termwise in that of $\mathfrak{g}$, so $\mathfrak{h}$ is solvable. Step 1.2 gives $0\mathrm{\ne }v\in V$ and $\lambda \in {\mathfrak{h}}^{\ast }$ with $hv=\lambda (h)v$ for every $h\in \mathfrak{h}$.

Put $W_j=\mathrm{span}(v,xv,\dots ,x^{j}v)$ and $W={\sum }_{j\ge 0}{W}_j$; finite dimensionality makes $W$ finite-dimensional and $x$-stable. Induction on $j$, using $h{x}^j=x^{j}h+{\sum }_{i=0}^{j-1}{x}^i[h,x]x^{j-1-i}$ and $[h,x]\in \mathfrak{h}$, shows $h{W}_j\subseteq W_j$ and $h{x}^{j}v\equiv \lambda (h)x^{j}v(\mathrm{m}\mathrm{o}\mathrm{d}{W}_{j-1})$. Thus $W$ is $\mathfrak{g}$-stable and every $h\in \mathfrak{h}$ acts upper triangularly on a basis extracted from the cyclic list, with constant diagonal $\lambda (h)$.

Let $d=\dim W>0$. Because $W$ is stable, [L2] and [L4] give $0={\mathrm{tr}}_W([x,h])=d\lambda ([x,h])$ for every $h\in \mathfrak{h}$, where the last equality uses the constant diagonal from step 3.1. Characteristic zero makes $d\cdot 1_k\mathrm{\ne }0$, so $\lambda ([x,h])=0$. This is the exact use of the characteristic hypothesis.

The nonzero common $\mathfrak{h}$-weight space $V_{\lambda }=\{u:hu=\lambda (h)u\text{ for all }h\in \mathfrak{h}\}$ contains $v$. It is $x$-stable: for $u\in V_{\lambda }$, [L2] and step 4.1 give $h(xu)=x(hu)+[h,x]u=\lambda (h)xu+\lambda ([h,x])u=\lambda (h)xu$.

By algebraic closure and [L3], $x{\mathrm{\mid }}_{V_{\lambda }}$ has a nonzero eigenvector $w$, say $xw=aw$. Then $hw=\lambda (h)w$ for $h\in \mathfrak{h}$, and for $y=h+tx$ we have $yw=(\lambda (h)+ta)w$. This defines the required linear functional $\chi$. Algebraic closure is used only in [L3], characteristic zero only in step 4.1, and all choices are a finite sequence of existential choices rather than AC.

If $V=0$, the empty flag and empty basis have the required properties.

Assume $V\mathrm{\ne }0$ and the corollary for smaller-dimensional modules.

By [L1], choose a common eigenvector $0\mathrm{\ne }v\in V$. Its line $V_1=kv$ is $\mathfrak{g}$-invariant.

The quotient $V\mathrm{/}{V}_1$ has the induced representation by [L2] and dimension one less. Step 1.2 supplies its complete invariant flag. Taking inverse images under the projection in [L3] and adjoining $0\subset V_1$ gives a complete invariant flag in $V$.

A basis adapted to this finite flag makes every representing matrix upper triangular. Conversely, the spans of the first $i$ vectors in a common upper-triangular basis form the complete invariant flag. These are finite sequential basis choices and require no AC.

By [L1], choose a common eigenvector $0\mathrm{\ne }v\in V$. Its line $kv$ is a nonzero $\mathfrak{g}$-invariant subspace.

Irreducibility [L2] forces $kv=V$, so ${\dim }_{\mathbb{C}}V=1$. The zero module is excluded by the definition of irreducibility used here; if $\mathfrak{g}=0$, the same argument says an irreducible nonzero module is one-dimensional.

Suppose first that $n=\dim V>0$. By [L1], choose a basis of $V$ in which every element of $\mathfrak{g}$ is upper triangular. The diagonal of $AB-BA$ is zero for upper triangular $A,B$, so every bracket is strictly upper triangular. Since strictly upper triangular matrices form a linear subspace, every element of $[\mathfrak{g},\mathfrak{g}]$, which is a finite linear combination of brackets, is strictly upper triangular and therefore has its $n$th power zero.

Put $\mathfrak{h}=[\mathfrak{g},\mathfrak{g}]$. It is a finite-dimensional Lie subalgebra of $\mathfrak{g}\mathfrak{l}(V)$, and step 1.1 says that for each $x\in \mathfrak{h}$ some $m>0$ satisfies $x^m=0$. On $\mathrm{End}(V)$ one has ${\mathrm{ad}}_x=L_x-R_x$; the operators $L_x$ and $R_x$ commute, and every term of $(L_x-R_x{)}^{2m-1}$ contains either $L_x^m$ or $R_x^m$. Hence ${\mathrm{ad}}_x$ is nilpotent on $\mathrm{End}(V)$ and therefore on its invariant subspace $\mathfrak{h}$. Engel's theorem [L2] makes $\mathfrak{h}$ nilpotent. If $V=0$, then $\mathfrak{g}=0$ and both conclusions are immediate.

Choose one finite basis $e_1,\dots ,e_n$ of $\mathfrak{g}$ and let $k_0\subseteq k$ be the subfield generated over $\mathbb{Q}$ by the finitely many structure constants in $[e_i,e_j]={\sum }_s{c}_{ij}^s{e}_s$. The same table defines a $k_0$-Lie algebra ${\mathfrak{g}}_0$ with $k{\otimes }_{k_0}{\mathfrak{g}}_0\cong \mathfrak{g}$ as in [L5]. Direct expansion of pure tensors and induction give $(K{\otimes }_F\mathfrak{a}{)}^{(r)}=K{\otimes }_F{\mathfrak{a}}^{(r)}$ and ${\gamma }_r(K{\otimes }_F\mathfrak{a})=K{\otimes }_F{\gamma }_r(\mathfrak{a})$ for every field extension $K\mathrm{/}F$. Scalar extension is faithful on a finite-dimensional vector space because a basis remains a basis. Hence solvability of $\mathfrak{g}$ implies solvability of ${\mathfrak{g}}_0$.

The finitely generated field $k_0\mathrm{/}\mathbb{Q}$ is countable with an explicit enumeration by rational expressions. In ZF, build an algebraic closure $K$ by a deterministic countable tower: dovetail all polynomials over earlier stages, choose the first coded monic irreducible factor, adjoin one root, and take the union. Every polynomial over the union occurs at a finite stage and later gains a root, so the union is algebraically closed. Put $\mathfrak{G}=K{\otimes }_{k_0}{\mathfrak{g}}_0$; step 1.1 makes $\mathfrak{G}$ finite-dimensional and solvable. No choice function is used in this fixed enumeration.

The homomorphic image $\mathrm{ad}(\mathfrak{G})$ is solvable because its derived terms are images of those of $\mathfrak{G}$. It is a linear Lie algebra on $\mathfrak{G}$, so [L1] says its derived algebra $[\mathrm{ad}(\mathfrak{G}),\mathrm{ad}(\mathfrak{G})]=\mathrm{ad}({\mathfrak{G}}^{\mathrm{\prime }})$ is nilpotent.