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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?

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  • $\begingroup$ 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. $\endgroup$ – Torsten Schoeneberg Nov 25 '18 at 19:44
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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.

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

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