Recall that a group $G$ satisfies the normalizer condition if for any proper subgroup $H$, its normalizer in $G$, $N_G(H)$ is a strictly larger group.
For finite groups, this property is equivalent to $G$ being nilpotent (that is, its lower central series terminates at the trivial group). The proof I know/found is by using yet another criteria for nilpotency: all Sylow subgroups are normal.
However, is there a proof that avoids mention of Sylow subgroups? I ask because both the condition of being nilpotent and having the normalizer condition are quite elementary and make no reference to Sylow subgroups.
Edit: I can at least show that the derived subgroup is a proper subgroup:
Let $M$ be a maximal subgroup of $G$, they exist by finiteness. The normalizer of $M$ properly contains $M$ and hence it is $G$, hence $M$ is normal in $G$. Moreover, $G/M$ has no subgroups and hence is prime cyclic. Now consider [G,G].
We will show that this is contained in $M$. This follows easily since [G/M,G/M] = [G,G]/M but $[G/M,G/M]$ is trivial since $G/M$ is abelian.
Best Answer
Personally, I don't believe that the theory of central series is more elementary than Sylow theory, and in my experience students find Sylow theory easier, probably because it does not involve technical calculations with quotient groups.
But if you insist, I think you can do it this way. By using induction on $|G|$, we can assume that $Z(G)=1$. Also all maximal subgroups $M$ of $G$ are normal, so they have prime index $p$. It is not hard to show that $M$ satisfies the normalizer condition, so $M$ is nilpotent by induction. So $Z(M) \ne 1$. Let $N$ be a minimal normal subgroup of $G$ that is contained in $Z(M)$, let $G = \langle M,g \rangle$ with $g^p \in M$, and let $H = \langle N,g \rangle$.
So $Z(H) \cap N \le Z(G)$ and hence $Z(H) \cap N = 1$. Since $\langle g \rangle \cap N \le Z(H) \cap N$, we have $\langle g \rangle \cap N =1$.
Now $H$ satifies the normalizer condition, by assumption if $H=G$, and by induction otherwise. So $\langle g \rangle < N_H(\langle g \rangle)$, and hence $N_N(\langle g \rangle) \ne 1$. But $[N_N(\langle g \rangle), \langle g \rangle] \le \langle g \rangle \cap N =1$, so $N_N(\langle g \rangle) \le Z(G)$, contradiction.