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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. 1 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 }. 2 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 3 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.) 4