6.5: Dihedral Groups
( \newcommand{\kernel}{\mathrm{null}\,}\)
Dihedral groups are groups of symmetries of regular n-gons. We start with an example.
Example 6.5.1
Consider a regular triangle T, with vertices labeled 1, 2, and 3. We show T below, also using dotted lines to indicate a vertical line of symmetry of T and a rotation of T.
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, and 2 and 3 mapping to each other. Similarly, if we rotate T clockwise by 120∘ (indicated in the picture by r), T again maps onto itself, this time with 1 mapping to 2, 2 mapping to 3, and 3 mapping to 1. Both of these maps are called symmetries of T; f is a reflection or flip and r is a rotation.
Of course, these are not the only symmetries of T. If we compose two symmetries of T, we obtain a symmetry of T: for instance, if we apply the map f∘r to T (meaning first do r, then do f) we obtain reflection over the line connecting 2 to the midpoint of line segment ¯13. Similarly, if we apply the map f∘(r∘r) to T (first do r twice, then do f) we obtain reflection over the line connecting 3 to the midpoint of line segment ¯12. In fact, every symmetry of T can be obtained by composing applications of f and applications of r.
For convenience of notation, we omit the composition symbols, writing, for instance, fr for f∘r, r∘r as r2, etc. It turns out there are exactly six symmetries of T, 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 ¯23);
- r (that is, clockwise rotation by 120∘);
- r2 (that is, clockwise rotation by 240∘);
- fr (that is, reflection over the line connecting 2 and the midpoint of ¯13); and
- fr2 (that is, reflection over the line connecting 3 and the midpoint of ¯12).
Declaring that f0=r0=e, the set
D3={e,f,r,r2,fr,fr2}={firj:i=0,1,j=0,1,2}
is the collection of all symmetries of T.
Remark
Notice that rf=fr2 and that f2=r3=e.
Theorem 6.5.1
D3 is a group under composition.
- Proof
-
First, as noted above, rf=fr2. So any map of the form firjfkrl(i,k=0,1,j,l=0,1,2) can be written in the form fsrt for some s,t∈N. Finally, let R2(s) and R3(t) be the remainders when you divide s by 2 and t by 3; then fsrt=fR2(s)rR3(t)∈D3. So ⟨D3,∘⟩ is a binary structure.
Next, function composition is always associative, and the function ee clearly acts as identity element in D3. Finally, let x=firj∈D3. Then y=r3−jf2−i is in D3 with xy=yx=e. So D3 is a group.
Let us look at D3 another way. Note that each map in D3 can be uniquely described by how it permutes the vertices 1,2,3 of T: that is, each map in D3 can be uniquely identified with a unique element of S3. For instance, f corresponds to the permutation (23) in S3, while fr corresponds to the permutation (13). In turns out that D3≃S3, via the following correspondence.
e↦e
f↦(23)
r↦(123)
r2↦(132)
fr↦(13)
fr2↦(12)
The group D3 is an example of class of groups called dihedral groups.
Definition: Dihedral Group
Let n be an integer greater than or equal to 3. We let Dn be the collection of symmetries of the regular n-gon. It turns out that Dn 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 D2n—so, for instance, our D3, above, would instead be called D6. Make sure you are aware of the convention your book or colleague is using.)
Theorem 6.5.2
Let n be an integer greater than or equal to 3. Then, again using the convention that f0=r0=e, Dn can be uniquely described as
Dn={firj:i=0,1,j=0,1,…,n−1}
with the relations
rf=frn−1 and f2=rn=e.
The dihedral group Dn is a nonabelian group of order 2n.
- Proof
-
The proof that Dn is a group parallels the proof, above, that D3 is a group. It is clear that Dn is nonabelian (e.g., rf=frn−1≠fr) and has order 2n.
Remark
Throughout this course, if we are discussing a group Dn you should assume n∈Z+, n≥3, unless otherwise noted.
Definition: Standard Form
We say that an element of Dn is written in standard form if it is written in the form firj where i∈{0,1} and j∈{0,1,…,n−1}.
Theorem 6.5.3
Each Dn is isomorphic to a subgroup of Sn.
- Proof
-
We provide here a sketch of a proof; the details are left as an exercise for the reader. We described above how D3 is isomorphic to a subgroup (namely, the improper subgroup) of S3. One can show that each Dn is isomorphic to a subgroup of Sn by similarly labeling the vertices of the regular n-gon 1,2,…,n and determining how these vertices are permuted by each element of Dn.
Note
While D3 is actually isomorphic to S3 itself, for n>3 we have that Dn is not isomorphic to Sn but is rather isomorphic to a proper subgroup of Sn. When n>3 you can see that Dn cannot be isomorphic to Sn since |Dn|=2n<n!=|Sn| for n>3.
It is important to be able to do computations with specific elements of dihedral groups. We have the following theorem.
Theorem 6.5.4
The following relations hold in Dn, for every n:
-
For every i, rif=fr−i (in particular, rf=fr−1=frn−1);
-
o(fri)=2 for every i (in particular, f2=e);
-
o(r)=o(r−1)=n;
-
If n is even, then rn/2 commutes with every element of Dn.
- Proof
-
- We use induction on the exponent of r. We already know that r1f=fr−1. Now suppose ri−1f=fr−(i−1) for some i≥2. Then
rif=r(ri−1f)=r(fr−(i−1))=(rf)r−i+1=(fr−1)r−i+1=fr−i.
- For every i, fri≠e, but
(fri)2=(fri)(fri)=f(rif)ri=f(fr−i)ri=f2r0=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 6.5.2
- Write fr2f in D3 in standard form. Do the same for fr2f in D4.
- What is the inverse of fr3 in D5? Write it in standard form.
- Explicitly describe an isomorphism from D4 to a subgroup of S4.
Example 6.5.3
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.)
Z,Z6,Z2,S6,Z4,Q,3Z,R,S2,R∗,S3,Q∗,C∗,⟨π⟩ in R∗,
D6,⟨(134)(25)⟩ in S5,R+,D3,⟨r⟩ in D4,17Z