Let $L$ be a restricted Lie algebra over a field $F$ of characteristic $p>0$. Assume that the following condition holds:


For every restricted ideal $I$ of $L$, the minimal restricted subalgebras of $L/I$ are pairwise non-isomorphic.



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> **QUESTION**: Is $L$ necessarily abelian?
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I already know that the answer is affirmative if one assumes that $L$ is nilpotent.



A condition on minimal restricted subalgebras of a restricted Lie algebra
The answer is positive if $L$ is finite-dimensional and $F$ is algebraically closed.


Indeed, suppose first that $L$ is $p$-nilpotent. Then the Frattini restricted subalgebra of $L$ is given by $\Phi(L)=[L,L]+L^{[p]}$, and $L/\Phi(L)$ is abelian with trivial $p$-map. Thus the hypothesis forces that $L/\Phi(L)$ is 1-dimensional, so that $L$ is cyclic and we are done.


On the other hand, if $L$ has an element $x$ that is not $p$-nilpotent, then consider the Jordan-Chevalley decomposition $x=x\_n+x\_s$ of $x$, where $x\_n$ is $p$-nilpotent, $x\_s$ is semisimple, and $[x\_n,x\_s]=0$. Then the restricted subalgebra $T$ generated by $x\_s$ is a torus and so, as $F$ is algebraically closed, $T$ has an $F$-basis consisting of elements $t$ such that $t^{[p]}=t$. By the hypothesis, we deduce that $L$ contains a unique (1-dimensional) torus $T$. Using again the fact that $F$ is algebraically closed, it follows that $L$ is nilpotent and so $T$ is contained in the center $Z(L)$ of $L$. In particular, $L/Z(L)$ is $p$-nilpotent and from the first part we infer that $L/Z(L)$ is cyclic. This implies that $L$ is abelian, yielding the claim.


