Question

Determine values of n for which Dihedral group Dn is nilpotent.

Answer 1

We denote by $D_{2n}$ the dihedral group of order $2n$.

$D_{2n}$ is nilpotent if n a power of 2: $n=2^m$ with $m\geq0$. In fact,

Suppose $D_{2 n}$ is nilpotent. Let $p$ be prime number which divides $n$ . Then $r^{n / p}$ is an element of $D_{2 n}$ order $p$, so $r^{n / p} \neq r^{-n / p}$ . Let $|s|=2$ and $|r^{n / p}|=p$ are relatively prime, so that, $s r^{n / p}=r^{n / p} s$ ; a contradiction. Thus no primes divide $n$ , and we have $n=2^{k}$.

reciprocally, we proceed by induction on $k$ where $n=2^{k}$.

For $k=0$, $D_{2.2^{0}} \cong Z_{2}$ is abelian, hence nilpotent.

Suppose $D_{2.2^{k}}$ is nilpotent. We have $Z\left(D_{2 \cdot 2^{k+1}}\right)=\left\langle r^{2^{k}}\right\rangle,$ and so, $D_{2 \cdot 2^{k+1}} / Z\left(D_{2 \cdot 2^{k+1}}\right) \cong D_{2 \cdot 2^{k}}$ is nilpotent. Thus, $D_{2 \cdot 2^{k+1}}$ is nilpotent.