suppose that $L$ is Lie(Leibniz) Algebra and $L^2$ is nilpotent.How to show that for any subalgebra $M$ of $L$, $\Phi(M)\subseteq \Phi(L)$?
$\Phi(L)$ is frattini ideal.
Thanks for help
Frattini and nilpotency
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lie-algebras
1 Answers
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This is proved in E.L. Stitzinger, Frattini subalgebras of a class of solvable Lie algebras, Pacific J. Math. 34 (1970) 177–182.