Theorem 5 (Levi–Malcev)

A Lie algebra $g$ always has a Levi subalgebra $s$. Every Levi subalgebra of $g$ is the image of $s$ under a special automorphism.

Let $r$ denote the radical of $g$. We first treat two special cases.

(a) $[g, r] = \{0\}$.

By Proposition 5, $g$ is then the product of its centre $r$ by $\mathscr{D}g$ which is semi-simple. Hence $\mathscr{D}g$ is a Levi subalgebra. Moreover, if $s'$ is a semi-simple subalgebra, then $s' = \mathscr{D}s'$ (Theorem 1), hence $s' \subset \mathscr{D}g$ and $\mathscr{D}g$ is the unique Levi subalgebra of $g$.

(b) $[g, r] \neq \{0\}$ and the only ideals of $g$ contained in $r$ are $\{0\}$ and $r$.

Then $[g, r] = r,\ [r, r] = \{0\}$ and the centre of $g$ is zero. Let $M$ (resp. $N$) be the subspace of $\mathscr{L}(g)$ consisting of the linear mappings of $g$ into $r$ whose restriction to $r$ is a homothety (resp. zero); $N$ is therefore of codimension 1 in $M$. For $m \in M$, let $\lambda(m)$ denote the ratio of the homothety of $r$ defined by $m$. Let $\sigma$ be the representation of $g$ on $\mathscr{L}(g)$ canonically derived from the adjoint representation; recall that $\sigma(x).u = [\mathrm{ad}_g x, u]$ for all $x \in g$ and all $u \in \mathscr{L}(g)$.

Clearly $\sigma(x)(M) \subset N$ for all $x \in g$. Moreover, if $x \in r, y \in g$ and $u \in M$, then

since $[r, r] = \{0\}$; and (4) can be written:

As the centre of $g$ is zero, the mapping $x \mapsto \mathrm{ad}_g x$ defines a bijection $\phi$ of $r$ onto a subspace $P$ of $\mathscr{L}(g)$. This subspace is stable under $\sigma(g)$ and contained in $N$ since $r$ is a commutative ideal and (5) shows that $\sigma(x)(M) \subset P$ for $x \in r$. The representation of $g$ on $M/P = V$ derived from $\sigma$ is therefore zero on $r$ and defines a representation $\sigma'$ of the semi-simple algebra $g/r$ on $V$. For all $y \in g/r$, the space $\sigma'(y)(V)$ is contained in $N/P$, which is of codimension 1 in $V$. Consequently (no. 2, Lemma 3) there exists $u_0 \in M$ such that $\lambda(u_0) = -1$ and $\sigma(x).u_0 \in P$ for all $x \in g$. The mapping $x \mapsto \phi^{-1}(\sigma(x).u_0)$ is a linear mapping of $g$ into $r$. By (5) its restriction to $r$ is the identity mapping of $r$. Hence its kernel is a subspace $s$ of $g$ supplementary to $r$ in $g$. As $s$ is the set of $x \in g$ such that $\sigma(x).u_0 = 0$, $s$ is a subalgebra of $g$ and therefore a Levi subalgebra of $g$.

Let $s'$ be another Levi subalgebra. For all $x \in s'$, let $h(x)$ be the unique element of $r$ such that $x + h(x) \in s$. Since $s$ is a subalgebra and $r$ is commutative, for $x, y$ in $s'$:

hence By Remark 2 of no. 2, there exists $a \in r$ such that $h(x) = -[x, a]$ for all $x \in s'$. Then: As $r$ is commutative, $(\mathrm{ad}\, a)^2 = 0$ and hence $1 + \mathrm{ad}\, a = e^{\mathrm{ad}\, a}$. As $r = [g, r]$, $e^{\mathrm{ad}\, a}$ is a special automorphism of $g$. By (6) this special automorphism maps $s'$ to $s$.

(c) General case:

We argue by induction on the dimension $n$ of the radical. There is nothing to prove if $n = 0$ and hence it can be assumed that the theorem holds for Lie algebras whose radical is of dimension $< \dim r$. By (a) it suffices to consider the case where $[g, r] \neq \{0\}$. As $[g, r]$ is nilpotent (no. 4, Proposition 6), its centre $c$ is $\neq \{0\}$. Let $m$ be a minimal non-zero ideal of $g$ contained in $c$. If $m = r$, we have case (b). Therefore let $m \neq r$ and let $f$ be the canonical mapping of $g$ onto $g' = g/m$. The radical of $g'$ is $r' = r/m$. By the induction hypothesis, $g'$ has a Levi subalgebra $l'$. Then $b = f(l')$ is a subalgebra of g containing m such that h/m = h' is semi-simple and hence having m as radical. By the induction hypothesis h = m + s where s is a semi-simple subalgebra. Then the equality g' = r' + h' implies g = r + h = r + m + s = r + s and hence s is a Levi subalgebra of g.

Let s' be another Levi subalgebra of g. Then f(s) and f(s') are two Levi subalgebras of g' and there exists by the induction hypothesis a' ∈ [g', r'] such that e^{ad\alpha'}(f(s')) = f(s). If a ∈ [g, r] is such that f(a) = a', it follows that:

s_1 = e^{ada}(s') \subset m + s = h.

Then s_1 and s are two Levi subalgebras of h and by the induction hypothesis there exists b ∈ m such that e^{adb}(s_1) = s. Hence s = e^{adb}.e^{ada}(s'). Finally, as m is in the centre of [g, r], e^{adb}.e^{ada} = e^{ad(b+a)} and b + a ∈ [g, r], which completes the proof.