suppose that $L$ is nilpotent leibniz algebra.why $J(L)=\Phi(L)$.
$J(L)$ is jacobson radical.and $\Phi(L)$is frattini ideal of $L$.and we know that if $L$ is nilpotent $L^2=\Phi(L)$.
thanks your help
suppose that $L$ is nilpotent leibniz algebra.why $J(L)=\Phi(L)$.
$J(L)$ is jacobson radical.and $\Phi(L)$is frattini ideal of $L$.and we know that if $L$ is nilpotent $L^2=\Phi(L)$.
thanks your help
In a nilpotent Leibniz algebra L every maximal subalgebra of L is an ideal of L (see D.W.Barnes, Some theorems on Leibniz algebras, Comm. In Alg. 39. , 2011, 2463-2472).