Dihedral Groups

The dihedral group $D_{2n}$ is the group of symmetries of a regular $n$-gon.

Definition. A dihedral group $D_{2n}$ is generated by two elements $r$ and $s$, in which the order of $r$ is $n$ and the order of $s$ is $2$, and $sr = r^{-1}s$.

Property. The order of $D_{2n}$ is $2n$.

Proof.

The elements of $D_{2n}$ are precisely

\[D_{2n} = \set{e, r, r^2, ..., r^{n-1}, s, rs, r^{2}s, ..., r^{n-1}s}\]

To complete the definiton of $D_{2n}$, it is necessary to show that any two groups with the above property are isomorphic.

Theorem. Any two dihedral groups of the same order are isomorphic.

Proof.

Any element in $G$ can be expressed as a combination of $r$, $r^{-1}$, $s$ and $s^{-1}$, e.g. $ssr^{-1}s^{-1}rrrs$. As $s^{-1} = s$ and $r^{-1} = r^{n-1}$, we can rewrite them only in terms of $s$ and $r$. Also, for $g = xsry$, we can rewrite it as $xr^{n-1}sy$ and repeating this process we can express $g = r^{p}s^{q}$ for non-negative integers $p, q$. As $G$ is of order $2n$ and $s^2 = r^n = e$, we can always relabel $G$ as $\set{r^{p}s^{q} : 0 \le p < n, q = 0, 1}$. Hence, all groups of this form are isomorphic to each other.

Symmetries in Complex Plane

Let $P$ be a regular polygon with $n$ sides and $D_{2n}$ be the symmetries of $P$. By translating, rotating and scaling $P$ on the complex plane, we can assume the set $V$ of vertices of $P$ is the set of $n$-th roots of unity, i.e.

\[V = \set{1, \omega, \omega^2, ..., \omega^{n-1}}\]

where $\omega = \exp(2\pi i/n)$.

A symmetry of $P$ is an isometry of $\mathbb{C}$ that maps the vertices $V$ to itself. In order to leave the set unchanged then it has to fix the origin $0$ as it is the only point that is distance one from each of the vertices. As any isometry of $\mathbb{C}$ is of the form $z \mapsto az + b$ or a reflection $z \mapsto az^{\ast} + b$, it follows that the symmetry is either a rotation $z \mapsto az$ or a reflection $z \mapsto az^{\ast}$ in some line through the origin, with $|a| = 1$.

There are exactly two maps that map $1 \in V$ to a chosen point $\omega^k \in V$, namely $z \mapsto \omega^kz$ and $z \mapsto \omega^kz^{\ast}$, that suggests the order of $D_{2n}$ is $2n$.

Now let $R_n$ be the subgroup of symmetries of $P$ that are rotations, then $R_n$ is cyclic group of order $n$ and generated by $r$, where $r(z) = \omega z$ and $r^n = I$. Next let $s(z) = z^{\ast}$, then $s$ is in $D_{2n}$ but not $R_n$, so $D_2n$ has two disjoint cosets $R_n$ and $sR_n$. Similarily, $R_n$ and $R_ns$ are also disjoint cosets and

\[sR_n \cup R_n = D_{2n} = R_n \cup R_ns\]

Thus, $sR_n = R_ns$ and $sr = r^{k}s$ for some $k$. As $r(z) = \omega z$ and $s(z) = z^{\ast}$, we can derive that

\[sr = r^{-1}s = r^{n-1}s\]

which is the special relation of $D_{2n}$

To conclude, the dihedral group $D_{2n}$ is the group of order $2n$ that is generated by an element $r$ or order $n$ and an element $s$ of order $2$, where $sr = r^{-1}s$.

Alternative generators

Proposition. $D_{2n}$ can be generated by the two elements $rs$ and $s$, both of order $2$.

Proof.

To generate the group, let $x = rs$ and $y = s$ be the two elements of order $2$. We have $xy = rss = r$, so the rotations are $(xy)^k$ and reflections are $(xy)^ky = (xy)^{k-1}x$.

Proposition. If $G$ is a finite group that is generated by two distinct elements of order $2$, then $G$ is a dihedral group. (Converse of the above)

Proof.

Let $a, b$ be the distinct generators of $G$ where $a^2 = b^2 = e$. Let $r = ab$ and $s = a$, we have $sr = a(ab) = b = (ba)a = r^{-1}s$. Also, as $G$ is finite, $r$ has finite order $n$.

Base on the above, we have $G = \set{r^ps^q : 0 \le p < n, q = 0, 1}$, in total $2n$ elements. If two elements are the same, say $g_1 = r^{p_1}s^{q_1} = r^{p_2}s^{q_2} = g_2$, we have $r^{p_2 - p_1} = s^{q_2 - q_1}$, where $0 \le p’ = p_2 - p_1 < n$ and $q’ = q_2 - q_1 = 0, 1$. Suppose $q’ = 1$, we have $(ab)^{p’} = a$ and $aba…aba…aba = e$. We can keep eliminating the $a$’s and $b$’s in the surrounding and have $b = e$ which contradicts with $|b| = 2$, so $q’ = 0$. Thus, $r^{p’} = e$ as well and we have $p_2 = p_1$ and $q_2 = q_1$. It means that the $2n$ elements in $G$ are all distinct to each other and therefore $G$ has order $2n$.

Combining the above, $G$ matches the definition of dihedral group.

Properties

Property. The reflections $sr^k \in D_{2n}$ has order $2$.

Proof.

\[(sr^k)(sr^k) = (sr^k)(r^{n-k}s) = e\]

Relationship with cyclic groups

As $r$ has order $n$ and $s$ has order $2$, it looks like $D_{2n} \cong C_n \times C_2$. However, it is not true as $D_{2n}$ is not abelian.

Klein Four-group

Definition. The Klein four-group, denoted by $V_4$, is a group with four elements, in which all of them is of order $2$.

Proposition. $V_4 \cong D_4 \cong C_2 \times C_2$.

Proof.

$V_4$ is abelian as all elements have order $2$. For $a, b \in V_4$, $a^2 = b^2 = e$ and $ab = ba = b^{-1}a$. So, $V_4 \cong D_4$.

For $a, b, c \in V_4$,

  • $\langle a \rangle = \set{e, a}$ and $\langle b \rangle = \set{e, b}$ so $\langle a \rangle \cup \langle b \rangle = \set{e}$

  • $ab = ba$

  • $c = ab$

Hence, by Direct Product Theorem, $V_4 \cong C_2 \times C_2$.

References