Are all abelian subgroups of a dihedral group cyclic?

Question

Are all abelian subgroups of a dihedral group cyclic?

Attempt: I have counter-examples for n=1,2 so I know that it isn't true for n<3. Is it true for n≥3? How do you know this?

Answer 1

For $n$ equal to a multiple of $4$, $n=4k$ with $k \ge 1$, the Dihedral group $D_n$ contains a copy of the non-cyclic abelian group $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$.

One can see this geometrically. Picture a regular $4k$-sided polygon $P$, centered at the origin of the plane, having one opposite pair of sides parallel to the $x$-axis, and another opposite pair of sides parallel to the $y$-axis. The symmetry group of $P$ is $D_{4k}$, and it consists of rotations about the origin through angles which are multiples of $2\pi/4k$ plus reflections across lines that bisect opposite side pairs or that connect opposite vertex pairs. Both the $x$ and $y$-axes bisect opposite side pairs, so reflections across those two axes are in the symmetry group, and they generate a subgroup isomorphic to $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$.

Answer 2

Let $G\le D_{2n}$ be a subgroup and $H\le G$ the subgroup of rotations in $G$. As $H\le C_n<D_{2n}$ because the subgroup $C_n$ comprises all rotations in $D_{2n}$, we know $H$ is cyclic, so assume $H<G$ is proper.

(I am using $D_{2n}$ to denote the dihedral group of order $2n$.)

Rotations preserve orientation and reflections reverse them, so composing two reflections yields a rotation and composing one of each yields a reflection. Pick any reflection $g\in G\setminus H$. Then multiplying by $g$ is a bijection $H\leftrightarrow G\setminus H$ so $[G:H]=2$. As $H$ is cyclic, write $H=\langle h\rangle$. Then $G=\langle g,h\rangle$ is also a dihedral group $D_{2m}$ where $m=|h|$. If $G$ is abelian, then $G$ must be the Klein four group as that is the only abelian dihedral group, and $g$ must have order $2$ in $C_n$.

If $n$ is odd then $D_{2n}$ has no noncyclic abelian subgroups, but if $n$ is even then $D_{2n}$ has a copy of the Klein four group $C_2\times C_2$ generated by the unique $180^\circ$ rotation and any flip.

Answer 3

There are two notations a dihedral group and $n\in\mathbb{N}$? Some use the notation $D_n=<r,s\,|\,s^2=r^n=1,\,rs=sr^{-1}>$ whereas others use the notation $D_{2n}=<r,s\,|\,s^2=r^n=1,\,rs=sr^{-1}>$. I'll use the second one (the group itself doesn't change).

I wonder what are your counter-examples. For $n=1$, we have $|D_2|=2$ thus $D_2\cong Z_2$ where $Z_2$ is the cyclic group of order $2$. Thus $D_2$ is cyclic and any subgroup of $D_2$ is either $1$ or $D_2$. In both cases, we get a cyclic abelian subgroup. But for $n=2$, you're right because $D_4\cong V_4$ the Klein $4$-group thus $D_4$ is abelian but not cyclic.

Now for $n\ge 3$, I will give you a hint: if $n$ is even prove that $Z(D_{2n})=<r^\frac{n}{2}>$ is the center of $D_{2n}$. Think about the reflexion $s$. Like that you'll be able to make a counter-example (a non cyclic subgroup must contain $s$ and a rotation). The statement is therefore false for even $n$. Now if $n$ is odd prove that $Z(D_{2n})=1$. As we said, a non-cyclic subgroup of $D_{2n}$ must have $s$ and a rotation (a power of $r$). Would such a group be abelian? You can prove that it isn't the case. Thus all abelian subgroups of $D_{2n}$ for odd $n$ are cyclic.