Let $R$ be a nilpotent algebra ($R^n = \{0\}$ for some $n \ge 1$) and $A$ be a subalgebra of $R$. I want to show that exist a finite chain of subalgebras {$R_i$ | $i = 0, 1, ..., m $}, $m \ge 1$, such that $$R_0 = \{0\} \lhd R_1 = A \lhd R_2 \lhd \ldots \lhd R_m = R$$ ($\lhd$ - Two-Sided ideal)
Update: As a finite chain we can take: $R \rhd R^{2} \rhd \ldots \rhd R^n =\{0\}$. Then just add A to this chain. So we get: $\{0\} \subseteq R^{n}+A=A \subseteq R^{n-1}+A \subseteq \ldots \subseteq R + A = R$. So the question is how to show that it is still a chain of ideals? Math induction?