# maximal subalgebras of nilpotent Lie algebra

I know that if every maximal subalgebra is an ideal, then L is nilpotent. Is every maximal subalgebra of a nilpotent Lie algebra an ideal?

• Have you tried induction over the length of the central series? Namely, for abelian $L$ the result is trivial, and it seems like one sees the main arguments already in the next step, where $[L,L]$ is central. – Torsten Schoeneberg Nov 25 '18 at 19:44

For any Lie algebra $$L$$, the lower central series is defined by $$C_0(L) := L$$, $$C_{n+1}(L):= [L, C_n(L)]$$. It is well-known that $$L$$ is nilpotent if and only if $$C_n(L) = 0$$ for sufficiently high $$n$$.

Now let $$M$$ be a maximal proper subalgebra of a Lie algebra $$L$$. Then $$M +C_1(L)$$ is also a subalgebra, so by maximality we have either $$M +C_1(L)=M$$ or $$M +C_1(L)=L$$. In the first case, $$M$$ is an ideal. To see that the second case does not occur in our situation, first show via induction:

Lemma: $$L=M +C_1(L) \Rightarrow C_n(L)=C_n(M) +C_{n+1}(L)$$ for all $$n$$.

Namely,

$$C_{n+1}(L)\\ =[L, C_n(L)]= [M+C_1(L), C_n(M) +C_{n+1}(L)] =\\ \underbrace{[M, C_n(M)]}_{C_{n+1}(M)} + \underbrace{[M, C_{n+1}(L)] +[C_1(L), C_n(M)] + [C_1(L), C_{n+1}(L)]}_{C_{n+2}(L)}$$

Now since $$L$$ is nilpotent, there is $$k$$ with $$C_k(L) \neq 0 = C_{k+1}(L)$$, hence $$C_k(L) =C_k(M)$$. Working backwards ($$k\to k-1\to...$$) with the formula in the lemma gives $$C_i(L) =C_i(M) \subseteq M$$ for all $$i$$, in particular $$L=M$$, contradiction to $$M$$ being a proper subalgebra.

• Thank you for your help. – Afsaneh Nov 26 '18 at 19:47