## Dih(2n) is nilpotent if and only if n is a power of 2

Prove that $D_{2n}$ is nilpotent if and only if $n$ is a power of 2.

$(\Rightarrow)$ Suppose $D_{2n}$ is nilpotent. Let $p$ be an odd prime dividing $n$. Then $r^{n/p}$ is an element of order $p$ in $D_{2n}$; in particular, $r^{n/p} \neq r^{-n/p}$. Now $|s| = 2$ and $|r^{n/p}| = p$ are relatively prime, so that, by §6.1 #9, $sr^{n/p} = r^{n/p}s$; a contradiction. Thus no odd primes divide $n$, and we have $n = 2^k$.

$(\Leftarrow)$ We proceed by induction on $k$, where $n = 2^k$.

For the base case, $k = 0$, note that $D_{2 \cdot 2^0} \cong Z_2$ is abelian, hence nilpotent.

For the inductive step, suppose $D_{2 \cdot 2^k}$ is nilpotent. Consider $D_{2 \cdot 2^{k+1}}$; we have $Z(D_{2 \cdot 2^{k+1}}) = \langle r^{2^k} \rangle$, and by §3.1 #34, $D_{2 \cdot 2^{k+1}}/Z(D_{2 \cdot 2^{k+1}}) \cong D_{2 \cdot 2^k}$ is nilpotent. By §6.1 #6, $D_{2 \cdot 2^{k+1}}$ is nilpotent.

Moreover, we show that for $k \geq 1$, $D_{2\cdot 2^k}$ is nilpotent of nilpotence class $k$, by induction on $k$.

For the base case, $k = 1$, $D_{2\cdot 2} \cong V_4$ is abelian, and thus of nilpotence class 1.

Suppose now that $D_{2 \cdot 2^k}$ is nilpotent of nilpotence class $k$. Since $D_{2 \cdot 2^k} \cong D_{2 \cdot 2^{k+1}}/Z(D_{2 \cdot 2^{k+1}})$, by a previous lemma, $D_{2 \cdot 2^{k+1}}$ is nilpotent of nilpotence class $k+1$.