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According to Humphreys, a Cartan sub-algebra of a Lie algebra $L$ is a nilpotent Lie sub-algebra whose normalizer is itself.

Look at analogous things in groups, or even in just finite groups.

If $G$ is a finite group, and if $H$ is a subgroup such that $H$ is nilpotent and is equal to its normalizer in $G$, what such subgroups are called? Have such subgroups studied? Is their existence in all finite groups studied/known?

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Nilpotent self-normalising subgroups are called Carter-subgroups. Carter (1961) proved that any finite solvable group has a Carter subgroup, and all its Carter subgroups are conjugate subgroups (and therefore isomorphic). If a group is not solvable it need not have any Carter subgroups: for example, the alternating group $A_5$ of order $60$ has no Carter subgroups.

For linear groups one could consider Borel subgroups, which are self-normalizing. They are solvable but not nilpotent. But then one could consider a maximal torus, which is "almost" self-normalizing.

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    I was thinking this, but $P$ is nilpotent - no problem; the normalizer of $P$ need not be nilpotent (simplest examples is $A_3$ in $S_3$)! I am concerning *$H$ which are nilpotent* **and** *are equal to their normalizers in $G$*. Instead, $S_2$ in $S_3$ is an example of the kind I am looking; so question is, does such subgroups $H$ always exists in every finite group? – 2017-02-24
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    No, there does not always exist such a group - the name is "Carter-subgroup", I found out. So I changed my answer. – 2017-02-24
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    You should correct what you wrote about standard parabolic subgroups: With one exception they are not even solvable. Borel will be solvable but not nilpotent. Each Cartan subgroup of GL(n,C) will be conjugate to the diagonal subgroup, a maximal torus. They are "almost" self-normalizing (the quotient N(T)/T will be the Weyl group). – 2017-02-24
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    @MoisheCohen Yes, thank you. Parabolic subgroups came from another discussion, I took them out. – 2017-02-24