Download Dihedral Handout

Abstract Algebra
Instructor: Mohamed Omar
Handout - Dihedral Groups
Jan 30
Math 171
Recall that the Dihedral group D2n consists the 2n symmetries of a regular n-gon. In
this handout we shall explore D2n under the operation of composition. This is indeed a
group because the composition of any two symmetries of a square is a symmetry of the
square, and the operation of undoing a geometric symmetry is itself a geometry symmetry.
We will work with the case n = 6. Below is a diagram of a pentagon with its vertices labeled
accordingly.
1. Recall the symmetries r and s in D12 where r is rotation by 2π
6 about the center of
the hexagon and s is reflection about the line through the vertex 1 and the center of
the hexagon. Draw the result of applying r to the above hexagon. Draw the result of
applying s to the above hexagon.
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2. Now consider the operation of composition on D12 , where if g, h ∈ D12 then gh is the
symmetry of the hexagon obtained by first applying h then applying g. Under this
operation, write down the result of applying rsr to hexagon above, and prove rsr = s.
3. Observe that s2 = r6 = 1 and indeed |s| = 2 and |r| = 6. Thus the following elements
are in D12
{1, r, r2 , r3 , r4 , r5 , s, sr, sr2 , sr3 , sr4 , sr5 }.
Observe that ri 6= rj for any i 6= j. One way to prove this is to just observe that
they are different symmetries. Another way is to use groups! Indeed, if ri = rj then
multiplying by (r−1 )j on the left we get e = ri−j which is impossible (why?) Use this
type of argument to justify:
(a) sri 6= srj if i 6= j
(b) ri 6= srj for any i, j
As a consequence, all the elements in {1, r, r2 , r3 , r4 , r5 , s, sr, sr2 , sr3 , sr4 , sr5 } are distinct. Since D12 has size 12 (as we discussed last class), we conclude
D12 = {1, r, r2 , r3 , r4 , r5 , s, sr, sr2 , sr3 , sr4 , sr5 }.
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4. How does the group operation in D12 work? To figure this out, simplify the following:
(a) r4 · r5
(b) r4 · sr3 (Hint: Remember rsr = s)
(c) sr2 · sr5
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5. Establish the following properties for D2n for general n.
(a) Each of 1, r, r2 , . . . , rn−1 are distinct, and rn = 1, so |r| = n.
(b) |s| = 2
(c) sri 6= srj for 0 ≤ i 6= j ≤ n − 1.
(d) D2n = {1, r, r2 , . . . , rn−1 , s, sr, sr2 , . . . , srn−1 }.
(e) ri s = sr−i for 0 ≤ i ≤ n. (Hint: Recall rsr = s.)
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