Dihedral groups are groups of symmetries of regular \(n\)-gons. We start with an example.

Example6.5.1

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 \(\overline{12}\)).

First, as noted above, \(rf=fr^2\text{.}\) So any map of the form \(f^ir^jf^kr^l\) (\(i,k=0,1\text{,}\) \(j,l=0,1,2\)) can be written in the form \(f^sr^t\) for some \(s,t \in \N\text{.}\) Finally, let \(R_2(s)\) and \(R_3(t)\) be the remainders when you divide \(s\) by \(2\) and \(t\) by \(3\text{;}\) then \(f^sr^t=f^{R_2(s)}r^{R_3(t)} \in D_3\text{.}\) So \(\langle D_3,\circ\rangle \) is a binary structure.

Next, function composition is always associative, and the function \(e\) clearly acts as identity element in \(D_3\text{.}\) Finally, let \(x=f^ir^j\in D_3\text{.}\) Then \(y=r^{3-j}f^{2-i}\) is in \(D_3\) with \(xy=yx=e\text{.}\) 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.

Definition6.5.4

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.)

Theorem6.5.5

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

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., \(rf=fr^{n-1}\neq fr\)) and has order \(2n\text{.}\)

Remark6.5.6

Throughout this course, if we are discussing a group \(D_n\) you should assume \(n\in \Z^+\text{,}\) \(n\geq 3\text{,}\) unless otherwise noted.

Definition6.5.7

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 \(j\in \{0,1,\ldots,n-1\}\text{.}\)

Theorem6.5.8

Each \(D_n\) is isomorphic to a subgroup of \(S_n\text{.}\)

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 \(S_3\text{.}\) 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 \(1,2,\ldots, n\) and determining how these vertices are permuted by each element of \(D_n\text{.}\)

Warning6.5.9

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.

Theorem6.5.10

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 \(D_n\text{.}\)

We use induction on the exponent of \(r\text{.}\) We already know that \(r^1f=fr^{-1}\text{.}\) Now suppose \(r^{i-1}f=fr^{-(i-1)}\) for some \(i\geq 2\text{.}\) Then