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So I am just making my way into the theory of Lie Algebras. The question at hand comes from page 14 of Humphreys, Introduction to Lie Algebra and Representation Theory.

Given a finite dimensional nilpotent Lie algebra $L$, and a proper sub-algebra $K$ of $L$, prove that the normalizer of $K$ properly contains $K$.

I keep trying to use Engel's theorem, but I have only been able to show there is a vector such that $[v\ K]$ is in the center, which much be in $K$. Though the condition for $v$ was only that it is in $L$.

A proof or anything else would be helpful. Thank you.

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    $\begingroup$ Here is one way: Split into two cases. Note that the center of $L$ will normalize $K$, so if $K$ does not contain the center, you are done. If $K$ does contain the center, it corresponds to a subalgebra of $L/Z(L)$ which has strictly smaller dimension than $L$. Induction. $\endgroup$ – Tobias Kildetoft May 15 '13 at 17:36
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    $\begingroup$ Ah, the untangeable bravdo of induction. Thank you, I hadn't thought of that. I wonder, does the proposition hold in infinite dimensional spaces? $\endgroup$ – Pax Kivimae May 15 '13 at 17:40
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    $\begingroup$ I am not actually sure in the infinite-dimensional case. My guess would be that it still holds, but I am not sure. $\endgroup$ – Tobias Kildetoft May 15 '13 at 17:49
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    $\begingroup$ @TobiasKildetoft Won't your proof work for infinite dimensional Lie algebras if you recast it as an induction on the ascending central series (which, if I correctly understand the definitions, still has to be finite in a nilpotent algebra), instead of an induction on dimension? $\endgroup$ – Andreas Blass May 15 '13 at 21:03
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    $\begingroup$ I think the last name of the cited book's author is not "Humphrey" but "Humphreys". $\endgroup$ – Andreas Blass May 15 '13 at 21:04
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Here is my comment as an answer:

We do this by induction on the length of the ascending central series (which is finite since $L$ is nilpotent).

If $K$ does not contain $Z(L)$ then we are done, as clearly $Z(L)$ normalizes $K$. So we assume that $Z(L)\subseteq K$ and look at the quotient $L/Z(L)$ which is nilpotent and has a shorter ascending central series than $L$.

$K$ corresponds to a subalgebra $\overline{K}$ of $L/Z(L)$ since $Z(L)\subseteq K$, and by induction, the normalizer of $\overline{K}$ strictly contains $\overline{K}$. But anything in the preimage of this normalizer in $L$ will normalize $K$, and we are done.

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