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.