Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2015 / Paper 2 / 10 / a

Table of contents
- Solution a

Solution

The table describes the symmetry group of a regular hexagon. Choose

R

as a counterclockwise rotation through

π /3

and

m_{1}

as reflection in one symmetry axis. With

s = m_{1}

, the relations are

R^{6} = I, s^{2} = I, s R s = R^{- 1} .

The rotation entries already are

I, R, \dots, R^{5}

, and the table gives

R m_{j} = m_{j + 1}

with indices modulo six. Therefore

m_{j} = R^{j - 1} s (j = 1, \dots, 6), G = {R^{k}, R^{k} s : 0 \leq k < 6} .

The source writes

D_{6}

with the subscript counting polygon vertices; this dihedral group has order 12.

 Ancestors (12)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version