False statement: finite nilpotent groups and finite solvable groups are the same

Given: The hypotheses and objects in the false claim.

Every nilpotent group is solvable. Consequently every finite $p$-group is solvable. (Nilpotent groups, and in particular finite $p$-groups, are solvable).

The derived series of a group $G$ is defined recursively by $G^{(0)}=G,G^{(r+1)}=[G^{(r)},G^{(r)}]\mathrm{.}$ Each term is characteristic, hence normal, in the preceding term by thm-derived-subgroup-is-characteristic-and-abelianization-is-universal. (The derived series, solvable groups, and derived length).

Let $n\in \mathbb{N}$, so that $n=\{0,1,\dots ,n-1\}$ (def-natural-numbers). The symmetric group on $n$ letters is $S_n:=\mathrm{Sym}(n)=\mathrm{Sym}(\{0,1,\dots ,n-1\}),$ the group of all bijections of $n$ under composition (def-symmetric-group), with the composition convention. (The finite symmetric group $S_n$, one-line notation, and cycle notation).

For a finite group $G$, the following are equivalent: $G$ is nilpotent; every Sylow subgroup is normal; $G$ is the internal direct product of its Sylow subgroups; and every maximal subgroup of $G$ is normal. (Sylow and maximal-subgroup characterizations of finite nilpotence).