# Is the dihedral group $D_n$ nilpotent? solvable?

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?

• 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 ... Jun 15, 2014 at 15:55
• ... with the basic structural properties of the dihedral group.) Regards, Jun 15, 2014 at 15:57

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 $$n>0$$.

• Could the downvote be explained?
– user122283
Sep 2, 2015 at 15:53
• 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... 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.