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