Solution

Label the triangle's vertices counterclockwise as

1, 2, 3

. Let

r

rotate it by

2 π /3

and let

s

reflect it in the axis through vertex 1 and the midpoint of the opposite side. Then

r^{3} = e, s^{2} = e, srs = r^{- 1} .

The six dihedral group elements are

D_{3} = {e, r, r^{2}, s, rs, r^{2} s} .

The first three are the identity and the rotations by

12 0^{\circ}, 24 0^{\circ}

; the last three are reflections fixing vertices

1, 3, 2

, respectively. These are all possible symmetries: the image of one vertex has three choices, and the order of the other two has two choices.

Use rightmost-first composition. From the generator relation,

(r^{a} s^{b}) (r^{c} s^{d}) = r^{a + (- 1)^{b} c} s^{b + d},

with exponents reduced modulo three and two. The row-times-column multiplication table is

e r r^{2} s rs r^{2} s e e r r^{2} s rs r^{2} s r r r^{2} e r^{2} s s rs r^{2} r^{2} e r rs r^{2} s s s s rs r^{2} s e r r^{2} rs rs r^{2} s s r^{2} e r r^{2} s r^{2} s s rs r r^{2} e

Thus every symmetry is expressed in the two group generators

r, s

 Ancestors (13)

There are no discussions about this article yet.

There are currently no matching articles.