you should post questions like this in the algebra section if you want to get a fater response.

anyway, suppose first that every

-dimesnional subalgebra of

is abelian and let

we only need to show that

is nilpotent because then we will be done by Engel's theoem. so suppose that

is an eigenvector of

. then there exists and

such that

but then the subalgebra generated by

would be at most

-dimensional and thus, by our hypothesis, abelian (note that 1-dimensional algebras are always abelian). hence

which implies

so every eigenvalue of

is zero and therefore

is nilpotent.

conversely, suppose that

is nilpotent. so

for some integer

now, let

be a subalgebra of

and

suppose that

is not abelian. then there exist

such that

for some

and we may assume that

so

and

and, if we continue this process, we will eventually get

because

but

and

and so

this contradiction proves that

is abelian.