TITLE: Nilpotent groups are solvable
QUESTION [8 upvotes]: I know this should be obvious but somehow I can't seem to figure it out and it annoys me!
My definition of nilpotent groups is the following:
A group $G$ is nilpotent if every subgroup of $G$ is subnormal in $G$, or equivalently if $U<N_G(U)$ for all $U<G$. And my definition of solvable groups is that a group $G$ is solvable if $U'\neq U$ for all subgroups $1\neq U \le G$, where $U'$ is the commutator subgroup of $U$.
My question is then: why are nilpotent groups solvable?

REPLY [3 votes]: Added: This follows exactly the same lines as Arturo's answer above.

From your definitions, it is perhaps easiest to work with the maximal subgroup point of view on nilpotence.
To begin with, let $M$ be a maximal proper subgroup of $G$, which is nilpotent.  Then (by your definition of nilpotence) we must have that $M$ is normal in $G$.
Thus $G/M$ is a group with no proper subgroups, hence is cyclic of prime order, and so in particular is abelian.  Thus $G$ has a non-trivial abelian quotient, and so $G'$ is a proper subgroup of $G$.

Now we are done, provided that we can show that each subgroup of a nilpotent group is again nilpotent.  This is not completely obvious using your definition, but is not too hard:
Suppose that $H$ is a subgroup of $G$, and that $U$ is a proper subgroup of $H$. By assumption $N_G(U)$ properly contains $U$.  I claim that $N_H(U)$ (which equals $H \cap N_G(U)$) properly contains $U$ also.  Indeed if not, i.e. if $N_H(U) = U$, then
$U =H \cap N_G(U)$.  Now consider $N_G(N_G(U))$: we see that $$H \cap N_G(N_G(U)) = N_H(H \cap N_G(U)) = N_H(U) = U.$$  Continuing, we find 
that $H \cap N_G(N_G( ... N_G(U) ...)) = U$.  But since $N_G$ of any proper subgroup
properly contains that subgroup, the iterated $N_G$s on the left hand side eventually reach $G$,
and so we see that $H\cap G = U$.  But $H \cap G = H$, and so $H = U$, contradicting our assumption that $U$ is a proper subgroup of $H$.
Thus indeed $N_H(U)$ properly contains $U$, and so $H$ is again nilpotent.