problem about nilpotent leibniz algebra

Question

I want to prove this statement

if $A$ is nilpotent leibniz algebra then $ H \subsetneq N_A(H)$. $H$ is a subalgebra of $A$ Can you help me how to show this.thanks

Answer 1

This follows from Engel's theorem for Leibniz algebras, in the same way as it follows for nilpotent Lie algebras. The proof is given in the paper On some basic properties of Leibniz algebras by V.V. Gorbatsevich; here Engel's theorem for Leibniz algebras is Proposition $2$; and the corollaries show that a proper $H$ is strictly contained in both right and left normalizers.

Edit: The proof of the first corollary to proposition $2$ seems to be not convincing, see the comments below. Perhaps I should not delete this post, although it has downvotes, so that other people can see it. Please refer to YCor's proof.

Answer 2

The result is false if $H=A$ so I assume $H\neq A$. Here is a proof, which works in an arbitrary nilpotent algebra $A$, that every proper subalgebra $H$ is strictly contained in its 2-sided normalizer $N(H)=\{v:vH\cup Hv\subset H\}$. (Under this generality $N(H)$ is not necessarily a subalgebra; it's contained in the left normalizer, which, in the Leibniz case, is a subalgebra).

Define $Z(A)$ as the nilcenter of $A$ (those $x$ such that $xy=yx=0$ for all $y$, I don't know if there's a standard name). This is a 2-sided ideal.

Iterating $Z(\cdot)$, we define the ascending (nil)central series of $A$ as usual: $0=A_0\subset A_1\subset\dots$, where $A_{i+1}/A_i$ is by definition the nilcenter of $H/A_i$. Nilpotent means that $A_i=A$ for large enough $i$.

Since $H\neq A$, there exists $i$, which we choose minimal, such that $H$ contains $A_i$ but not $A_{i+1}$. So by definition, $A_{i+1}A$ and $AA_{i+1}$ are both contained in $A_i$, and hence in $H$. So if $N=A_{i+1}+H$, then $N$ is a subalgebra and $NH\cup HN\subset H$. By choice of $i$, $H$ is properly contained in $N$.