Is it true that a dihedral group is nonabelian?

Question

Is it true that a dihedral group is nonabelian?

I'm not sure if the result is true. I checked it for some lower order and I think the result may correct.

But I failed to prove/disprove the result.

Answer 1

Yes, the dihedral groups $D_n$ are nonabelian for $n\ge 3$. It is generated by a rotation $r$ with $r^n=1$ and a reflection $s$ with $s^2=1$. However, you can easily check that a rotation and a reflection will not commute in general. We have $sr=r^{-1}s$ instead for $D_n$ with this presentation.

Answer 2

$D_3$, i.e, the dihedral group of a triangle is isomorphic to $S_3$ which is non-abelian. It can be shown that this is true for $n \geq 3$.

Remark: Some people denote the dihedral group by $D_{2n}$ which is based on the fact the order is $2n$, while some people denote it by $D_n$.

Answer 3

The dihedral groups for $n=1$ and $n=2$ are abelian; for $n\geq 3$, the dihedral groups are nonabelian (this is mentioned on Wikipedia).

Answer 4

Dihedral group of order $2n$ has the presentation

$$\langle x,y \mid x^n=y^2=e,yxy=x^{-1}\rangle$$

When $n=1$, we have $x=e$ and thus it is a group of order 2.

When $n=2$, we have $yxy=x$ and so $xy=yx$. Thus it is abelian.

But when $n\geq 3$, then $yxy=x^{n-1}$ and so $xy=yx^{n-1} \neq yx$. Thus it is not abelian.

Answer 5

My picture for $D_4$ shows $fr = r^{-1}f$ and not $rf$.
See Zev Chonoles's exquisite pictures at https://math.stackexchange.com/a/686175/53934 too.