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?

  • 9
    $\begingroup$ Dear user, One of the most basic facts about the dihedral group of order $2n$ is that it contains a normal and cyclic subgroup of rotations of order $n$ with index $2$. If you don't see straight away that this implies that the dihedral group is solvable, then it would probably be a good idea to review the relevant background to this question carefully, so that your intuition catches up with your formal understanding. (What I mean is that, while nilpotency may a little more subtle, for solvability it shouldn't be a matter of applying theorems, but rather just combining the definition ... $\endgroup$
    – Matt E
    Jun 15, 2014 at 15:55
  • $\begingroup$ ... with the basic structural properties of the dihedral group.) Regards, $\endgroup$
    – Matt E
    Jun 15, 2014 at 15:57

2 Answers 2


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 $n>1$, $D_{2n}/\langle a\rangle\cong\mathbb{Z}_2$ and $\langle a\rangle \cong \mathbb{Z}_n$. 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 $n>0$.

  • $\begingroup$ Could the downvote be explained? $\endgroup$
    – user122283
    Sep 2, 2015 at 15:53
  • 2
    $\begingroup$ I didn't downvote, but using the argument "$G$ is nilpotent and so is solvable" when $G$ is abelian, or to say that "extensions of solvable groups are solvable" when you've explicit construct of series with abelian quotients, is somewhat weird... $\endgroup$ Sep 29, 2015 at 10:12

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.


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