Let 
be a finite-dimensional Lie algebra over some field
. A subalgebra 
of 
is called a Cartan subalgebra if it is nilpotent
and equal to its normalizer, which is the set of those elements 
such that ![[x,h] subset h](/images/equations/CartanSubalgebra/Inline6.svg)
.
It follows from the definition that if 
is nilpotent, then 
itself is a Cartan subalgebra of 
. On the other hand, let 
be the Lie algebra of all endomorphisms
of 
(for some natural number 
), with ![[f,g]=f degreesg-g degreesf](/images/equations/CartanSubalgebra/Inline13.svg)
. Then the set of all endomorphisms 
of 
of the form 
is a Cartan subalgebra
of 
.
It can be proved that:
1. If 
is infinite, then 
has Cartan subalgebras.
2. If the characteristic of 
is equal to 
, then all Cartan subalgebras of 
have the same dimension.
3. If 
is algebraically closed and its characteristic
is equal to 0, then, given two Cartan subalgebras 
and 
of 
, there is an automorphism 
of 
such that 
.
4. If 
is semisimple and 
is an infinite field whose characteristic
is equal to 0, then all Cartan subalgebras of 
are Abelian.
Every Cartan subalgebra of a Lie algebra 
is a maximal nilpotent subalgebra of 
. However, a maximal nilpotent subalgebra of 
doesn't have to be a Cartan subalgebra. For instance, if 
is the Lie algebra of all endomorphisms
of 
with ![[f,g]=f degreesg-g degreesf](/images/equations/CartanSubalgebra/Inline38.svg)
and if 
is the subalgebra of all endomorphisms 
of the form 
, then 
is a maximal nilpotent subalgebra of 
, but not a Cartan subalgebra.
