Definition. The $i$-higher commutator subgroup $\mathcal{D}_{i}(G)$ is defined by $$\mathcal{D}_{1}(G) = G$$ and $$\mathcal{D}_{i+1}(G) = [\mathcal{D}_{i}(G),G].$$
A Course on Finite Groups. Rose.
A group $G$ is nilpotent if there is a positive integer $r$ such that $\mathcal{D}_{r+1}(G) = \{e\}$.
I want to use this definition to show that $D_{8}$ is nilpotent. But, doing the calculations is very hard work. I know that $D_{8}$ is nilpotent with lower central series $D_{8} \geq C_{2} \geq \{e\}$.
Is there a clever way to show that $D_{8}$ is nilpotent using this definition without needing so many calculations?
My approach: by lower central series, we know that $\mathcal{D}_{3}(D_{8}) = \{e\}$ ($\ast$). But $$\mathcal{D}_{2}(D_{8}) = [\mathcal{D}_{1}(D_{8}),D_{8}] = [D_{8},D_{8}] = D_{8}'.$$ We know that $D_{8}' = \langle r^{2} \rangle$ where $r$ is the generator of order $n$.
Is correct? If yes, $\langle r^{2} \rangle$ is isomorphic to which subgroup of $D_{8}$?
I'm not sure about ($\ast$). Seems intuitive, but this means nothing. If is correct, someone can clarify to me why?
One characterization of the commutator subgroup $[G,G]$ of a nonabelian group $G$ is that it is the smallest nontrivial normal subgroup of $G$ such that the quotient $G/[G,G]$ is abelian in the sense that if $H$ is normal in $G$ and $G/H$ is abelian, then $[G,G]\subset H$.
With this characterization in mind, and the observation that $D_8/\langle r^2\rangle \cong \Bbb Z/2\Bbb Z\times\Bbb Z/2\Bbb Z$ is abelian, we can conclude that $[D_8,D_8] = \langle r^2\rangle$ since the commutator must be contained in $\langle r^2\rangle$, but $\langle r^2\rangle$ is a subgroup of order $2$.