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I am wondering is it really necessary to use the adjoint version of Engel’s theorem to prove that if every 2 dimensional lie sub-algebra of a given Lie algebra L is abelian then L is nilpotent? I wonder about that because you can write every 2 dimensional Lie algebra as the span of x and y in L and then by the assumption the bracket is zero for every two elements that I could choose. What am I missing here?

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  • $\begingroup$ It's false when the field is not algebraically closed, see my answer. $\endgroup$ – YCor Jul 31 at 21:30
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I think you are missing, that two arbitrary elements in $L$ will not span a subalgebra of $L$. For example, if $L$ is the Heisenberg Lie algebra with basis $(x,y,z)$ and Lie brackets determined by $[x,y]=z$, then the Lie bracket of each two elements spanning a subalgebra is zero, but not in general. $x$ and $y$ have a non-trivial Lie bracket and they do not span a subalgebra of $L$ (because the product is not "closed").

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Proposition. Let $\mathfrak{g}$ be a finite-dimensional Lie algebra over an algebraically closed field $K$. Then $\mathfrak{g}$ is nilpotent iff every 2-dimensional subalgebra is abelian.

One direction is trivial (over an arbitrary field). For the nontrivial implication, suppose that $\mathfrak{g}$ is not nilpotent. By Engel's theorem (which holds over every field, see Jacobson), there exists $x\in\mathfrak{g}$ such that $\mathrm{ad}(x)$ is not nilpotent. Since the field is algebraically closed, this means that $\mathfrak{x}$ has a nonzero eigenvalue $\lambda$, say for en eigenvector $y$. Then $(x,y)$ span a non-abelian 2-dimensional subalgebra.

Remark. The above fails for every field $K$ that is not algebraically closed.

Indeed, choose an irreducible polynomial of degree $d\ge 2$. Let $u$ be the companion matrix of this polynomial. Then $u$ defines a semidirect product Lie algebra $K^d\rtimes_uK$ (of dimension $d+1\ge 3$). Then all its proper subalgebras are abelian (in particular, the 2-dimensional subalgebras are the 2-dimensional subspaces of the abelian ideal $K^d$).

Still there is a replacement, essentially following from the proof of the proposition. Say that an operator $u$ on $K^d$ is irreducible if $d\ge 1$ and $u$ preserves no subspace else than $\{0\}$ and $K^d$. So there exists such operators on $K^d$ if and only if there's an irreducible polynomial of degree $d$.

Proposition. Let $\mathfrak{g}$ be a finite-dimensional Lie algebra over a field field $K$. Then $\mathfrak{g}$ is nilpotent iff for any $d\ge 1$ it has no subalgebra of the form $K^d\rtimes_u K$ with $u$ nonzero irreducible operator.

For the real field, up to conjugation and nonzero scalar multiplication, the only possibilities for $u$, namely $\begin{pmatrix}1\end{pmatrix}$, $\begin{pmatrix}t & -1\\1 & t\end{pmatrix}$ for $t\ge 0$. So this yields

Proposition. Let $\mathfrak{g}$ be a finite-dimensional Lie algebra over $\mathbf{R}$. Then $\mathfrak{g}$ is nilpotent if and only if it has no subalgebra isomorphic to any of the following: the 2-dimensional non-abelian Lie algebra, and for any $t\ge 0$, the 3-dimensional Lie algebra with basis $(h,x,y)$ and brackets $$[h,x]=tx+y,\quad[h,y]=-x+ty,\quad [x,y]=0.$$

Finally, let me mention that the criterion is also false, over $\mathbf{C}$, in infinite dimension (finite dimension is too often implicit). Indeed the (finitely generated) Lie algebra with basis $(x_n)_{n\ge 1}$ and only nonzero brackets $[x_1,x_i]=x_{i+1}$ is residually nilpotent and hence its finite-dimensional subalgebras are nilpotent (actually they're even abelian, exercise). But it's clearly not nilpotent since $\mathrm{ad}(x_1)$ is not nilpotent. Non-nilpotent, locally nilpotent Lie algebras (e.g., a direct sum of nilpotent Lie algebras of unbounded nilpotency class) are also counterexamples.

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