Is the dihedral group $D_n$ nilpotent? solvable?

Question

Is the dihedral group $D_n$ nilpotent? solvable?

I'm trying to solve this problem but I've been trying to apply a couple of theorems but have been unsuccessful so far. Can anyone help me?

Answer 1

Theorem:$D_n$ is nilpotent iff $n=2^i$ for $i\geq0$.

The following is the proof as given here:

($\Rightarrow$) 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, $sr^{n/p} = r^{n/p}s$; 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 so, $D_{2 \cdot 2^{k+1}}/Z(D_{2 \cdot 2^{k+1}}) \cong D_{2 \cdot 2^k}$ is nilpotent. Thus, $D_{2 \cdot 2^{k+1}}$ is nilpotent.

Theorem: $D_{2n}$ is solvable for all $n\geq1$.

To prove this, I will use the above fact (see here).

When $n=1$, $D_{2n}\cong \mathbb{Z}_2$ which is nilpotent and thus solvable. When $n>1$, $D_{2n}/\langle r\rangle\cong\mathbb{Z}_2$ and $\langle r\rangle \cong \mathbb{Z}_n$. Both $\mathbb{Z}_n$ and $\mathbb{Z}_2$ are nilpotent and so they are both solvable. As extensions of solvable groups are solvable, $D_{2n}$ is solvable for all $n>0$.

Answer 2

Here's another approach to show that $D_{n}$ is nilpotent iff $n$ is a power of $2$:

Suppose $n\geq 3$ and denote $\textbf{Z}=\textbf{Z}(D_n)$ as the center of $D_n$. It is not difficult to see that $\textbf{Z}=1$ if $n$ is odd and that $\textbf{Z}=\langle r^{n/2}\rangle$ if $n$ is even. Here $r$ denotes de rotation of $D_{n}$ and $s$ the reflection of $D_n$. Now, if $n$ is even, we have that $D_n/\textbf{Z}\cong D_{n/2}$. This follows because the cosets $r\textbf{Z}$ and $s\textbf{Z}$ satisfy the presentation equations for $D_{n/2}$, that is, $$(r\textbf{Z})^{n/2}=(s\textbf{Z})^2=(r\textbf{Z}s\textbf{Z})^2=\textbf{Z}.$$ Thus, the upper central series of $D_n$ $$1\leq \textbf{Z}\leq Z_2\leq Z_3\leq \cdots$$ will reach $D_n$ iff the numbers $n$, $n/2$, $n/4$, ..., are all even.

Answer 3

A general proof that $D_n$ is solvable for all $n \geq 1$.

Here is the normal series with abelian quotients: $$\{e\} \trianglelefteq \langle \sigma \rangle \trianglelefteq D_n$$

$$\langle \sigma \rangle/\{e\} \cong \mathbb{Z}_n, D_n/\langle \sigma \rangle \cong \mathbb{Z}_2$$ both obviously abelian, therefore $D_n$ is solvable.