6.5: Dihedral Groups
Dihedral groups are groups of symmetries of regular \(n\)-gons. We start with an example.
Example \(\PageIndex{1}\)
Consider a regular triangle \(T\text{,}\) with vertices labeled \(1\text{,}\) \(2\text{,}\) and \(3\text{.}\) We show \(T\) below, also using dotted lines to indicate a vertical line of symmetry of \(T\) and a rotation of \(T\text{.}\)
Note that if we reflect \(T\) over the vertical dotted line (indicated in the picture by \(f\)), \(T\) maps onto itself, with \(1\) mapping to \(1\text{,}\) and \(2\) and \(3\) mapping to each other. Similarly, if we rotate \(T\) clockwise by \(120^{\circ}\) (indicated in the picture by \(r\)), \(T\) again maps onto itself, this time with \(1\) mapping to \(2\text{,}\) \(2\) mapping to \(3\text{,}\) and \(3\) mapping to \(1\text{.}\) Both of these maps are called symmetries of \(T\text{;}\) \(f\) is a reflection or flip and \(r\) is a rotation .
Of course, these are not the only symmetries of \(T\text{.}\) If we compose two symmetries of \(T\text{,}\) we obtain a symmetry of \(T\text{:}\) for instance, if we apply the map \(f\circ r\) to \(T\) (meaning first do \(r\text{,}\) then do \(f\)) we obtain reflection over the line connecting \(2\) to the midpoint of line segment \(\overline{13}\text{.}\) Similarly, if we apply the map \(f\circ (r\circ r)\) to \(T\) (first do \(r\) twice, then do \(f\)) we obtain reflection over the line connecting \(3\) to the midpoint of line segment \(\overline{12}\text{.}\) In fact, every symmetry of \(T\) can be obtained by composing applications of \(f\) and applications of \(r\text{.}\)
For convenience of notation, we omit the composition symbols, writing, for instance, \(fr\) for \(f\circ r\text{,}\) \(r\circ r\) as \(r^2\text{,}\) etc. It turns out there are exactly six symmetries of \(T\text{,}\) namely:
- the map \(e\) from \(T\) to \(T\) sending every element to itself;
- \(f\) (that is, reflection over the line connecting \(1\) and the midpoint of \(\overline{23}\));
- \(r\) (that is, clockwise rotation by \(120^{\circ}\));
- \(r^2\) (that is, clockwise rotation by \(240^{\circ}\));
- \(fr\) (that is, reflection over the line connecting \(2\) and the midpoint of \(\overline{13}\)); and
- \(fr^2\) (that is, reflection over the line connecting \(3\) and the midpoint of \(\overline{12}\)).
Declaring that \(f^0=r^0=e\text{,}\) the set
\begin{equation*} D_3=\{e, f, r, r^2, fr, fr^2\}=\{f^ir^j:i=0,1, j=0,1,2\} \end{equation*}
is the collection of all symmetries of \(T\text{.}\)
Remark
Notice that \(rf=fr^2\) and that \(f^2=r^3=e\text{.}\)
Theorem \(\PageIndex{1}\)
\(D_3\) is a group under composition.
- Proof
-
First, as noted above, \(rf=fr^2\). So any map of the form \(f^ir^jf^kr^l\)\((i,k=0,1,\;\; j,l=0,1,2)\) can be written in the form \(f^sr^t\) for some \(s,t∈N\). Finally, let \(R_2(s)\) and \(R_3(t)\) be the remainders when you divide \(s\) by \(2\) and \(t\) by \(3\); then \(f^sr^t=f^{R_2(s)}r^{R_3(t)}∈D_3\). So \(\langle D_3, \circ \rangle\) is a binary structure.
Next, function composition is always associative, and the function ee clearly acts as identity element in \(D_3\). Finally, let \(x=f^ir^j∈D_3\). Then \(y=r^{3−j}f^{2−i}\) is in \(D_3\) with \(xy=yx=e\). So \(D_3\) is a group.
Let us look at \(D_3\) another way. Note that each map in \(D_3\) can be uniquely described by how it permutes the vertices \(1,2,3\) of \(T\text{:}\) that is, each map in \(D_3\) can be uniquely identified with a unique element of \(S_3\text{.}\) For instance, \(f\) corresponds to the permutation \((23)\) in \(S_3\text{,}\) while \(fr\) corresponds to the permutation \((13)\text{.}\) In turns out that \(D_3 \simeq S_3\text{,}\) via the following correspondence.
\(e \mapsto e\)
\(f \mapsto (23)\)
\(r \mapsto (123)\)
\(r^2 \mapsto (132)\)
\(fr \mapsto (13)\)
\(fr^2 \mapsto (12)\)
The group \(D_3\) is an example of class of groups called dihedral groups .
Definition: Dihedral Group
Let \(n\) be an integer greater than or equal to \(3\text{.}\) We let \(D_n\) be the collection of symmetries of the regular \(n\)-gon. It turns out that \(D_n\) is a group (see below), called the dihedral group of order \(2n\). (Note: Some books and mathematicians instead denote the group of symmetries of the regular \(n\)-gon by \(D_{2n}\)—so, for instance, our \(D_3\text{,}\) above, would instead be called \(D_6\text{.}\) Make sure you are aware of the convention your book or colleague is using.)
Theorem \(\PageIndex{2}\)
Let \(n\) be an integer greater than or equal to \(3\text{.}\) Then, again using the convention that \(f^0=r^0=e\text{,}\) \(D_n\) can be uniquely described as
\begin{equation*} D_n=\{f^ir^j: i=0,1, j=0,1,\ldots, n-1\} \end{equation*}
with the relations
\begin{equation*} rf=fr^{n-1} \text{ and } f^2=r^n=e. \end{equation*}
The dihedral group \(D_n\) is a nonabelian group of order \(2n\text{.}\)
- Proof
-
The proof that \(D_n\) is a group parallels the proof, above, that \(D_3\) is a group. It is clear that \(D_n\) is nonabelian (e.g., \(rf=fr^{n−1}≠fr\)) and has order \(2n\).
Remark
Throughout this course, if we are discussing a group \(D_n\) you should assume \(n\in \mathbb{Z}^+\text{,}\) \(n\geq 3\text{,}\) unless otherwise noted.
Definition: Standard Form
We say that an element of \(D_n\) is written in standard form if it is written in the form \(f^ir^j\) where \(i\in \{0,1\}\) and \(j\in \{0,1,\ldots,n-1\}\text{.}\)
Theorem \(\PageIndex{3}\)
Each \(D_n\) is isomorphic to a subgroup of \(S_n\text{.}\)
- Proof
-
We provide here a sketch of a proof; the details are left as an exercise for the reader. We described above how \(D_3\) is isomorphic to a subgroup (namely, the improper subgroup) of \(S_3\). One can show that each \(D_n\) is isomorphic to a subgroup of \(S_n\) by similarly labeling the vertices of the regular \(n\)-gon \(1,2,…,n\) and determining how these vertices are permuted by each element of \(D_n\).
Note
While \(D_3\) is actually isomorphic to \(S_3\) itself, for \(n>3\) we have that \(D_n\) is not isomorphic to \(S_n\) but is rather isomorphic to a proper subgroup of \(S_n\text{.}\) When \(n>3\) you can see that \(D_n\) cannot be isomorphic to \(S_n\) since \(|D_n|=2n \lt n! = |S_n|\) for \(n>3\text{.}\)
It is important to be able to do computations with specific elements of dihedral groups. We have the following theorem.
Theorem \(\PageIndex{4}\)
The following relations hold in \(D_n\text{,}\) for every \(n\text{:}\)
-
For every \(i\text{,}\) \(r^if=fr^{-i}\) (in particular, \(rf=fr^{-1}=fr^{n-1}\));
-
\(o(fr^i)=2\) for every \(i\) (in particular, \(f^2=e\));
-
\(o(r)=o(r^{-1})=n\text{;}\)
-
If \(n\) is even, then \(r^{n/2}\) commutes with every element of \(D_n\text{.}\)
- Proof
-
- We use induction on the exponent of \(r\). We already know that \(r^1f=fr^{−1}\). Now suppose \(r^{i−1}f=fr^{−(i−1)}\) for some \(i≥2\). Then
\(r^if=r(r^{i−1}f)=r(fr^{−(i−1)})=(rf)r^{−i+1}=(fr^{−1})r^{−i+1}=fr^{−i}\).
- For every \(i\), \(fr^i≠e\), but
\((fr^i)^2=(fr^i)(fr^i)=f(r^if)r^i=f(fr^{−i})r^i=f^2r^0=e\).
- This follows from Theorem \(5.1.5\) and the fact that o(r)=n.o(r)=n.
- The proof of this statement is left as an exercise for the reader.
Example \(\PageIndex{2}\)
- Write \(fr^2f\) in \(D_3\) in standard form. Do the same for \(fr^2f\) in \(D_4\text{.}\)
- What is the inverse of \(fr^3\) in \(D_5\text{?}\) Write it in standard form.
- Explicitly describe an isomorphism from \(D_4\) to a subgroup of \(S_4\text{.}\)
Example \(\PageIndex{3}\)
Classify the following groups up to isomorphism. ( Hint: You may want to look at the number of group elements that have a specific finite order.)
\begin{equation*} \mathbb{Z}, \mathbb{Z}_6, \mathbb{Z}_2, S_6, \mathbb{Z}_4, \mathbb{Q}, 3\mathbb{Z}, \mathbb{R}, S_2, \mathbb{R}^*, S_3,\mathbb{Q}^*, \mathbb{C}^*, \langle \pi\rangle \text{ in } \mathbb{R}^*, \end{equation*}
\begin{equation*} D_6, \langle (134)(25)\rangle \text{ in } S_5, \mathbb{R}^+, D_3, \langle r \rangle \text{ in } D_4, 17\mathbb{Z} \end{equation*}