- [1] Let n > 2. An element of h of D
_{n}is*nontrivial*if it is not the identity. It is a*rotation*if (thinking of D_{n}as symmetries of a regular n-gon) it keeps the top side of the n-gon up. It is a*reflection*if it is nontrivial but some point on the perimeter of the n-gon is not moved. Determine whether the following statement is true or false. Justify it if it is true, or give a counterexample if it is false: Every nontrivial symmetry in D_{n}is either a reflection or a rotation.

We know |D_{n}| = 2n. We also know, in addition to the identity, there are n-1 rotations (a given vertex of the polygon can be rotated around to any of the other n-1 vertices) and there are n reflections (when n is odd these are the reflections across the perpendicular bisector of each side, and when n is even, these are the reflections across the n/2 angle bisectors and the n/2 perpendicular bisectors of the sides). Since this already gives us 2n elements, we see each nontrivial element of D_{n}is either a reflection or a rotation.

- [2]
- (a) Prove that the product of two rotations in D
_{n}is a rotation.

Composing two rotations keeps the same side of the polygon up, so it is by definition another rotation. (Of course, rotation through an angle b and then through an angle c just gives a rotation through an angle b + c.) Thus the composition of two rotations is a rotation.

- (b) Prove that the product of two reflections in D
_{n}is a rotation.

Composing two reflections results in turning the polygon over twice so the top side ends up facing up. Thus the composition is another rotation.

- (a) Prove that the product of two rotations in D
- [3] Consider an infinite strip of equally spaced H's in the plane:

... H H H H H H H H H H H H ... Describe the symmetries of the strip. Is the group of symmetries abelian?

The strip has four kinds of symmetries. You can slide it sideways by a multiple of the distance between two H's. You can reflect across the horizontal line though the center of the H's. You can reflect across vertical lines halfway between each adjacent pair of H's or through the center of an H. Finally, you can also rotate 180 degrees around a point midway between adjacent H's or at the center of an H (but this is the same as combining a reflection across a horizontal and then a vertical line). The group is not abelian. Suppose the strip of H's are arranged in the plane so that the horizontal bar in each H is on the x-axis, with the center point of each bar being at an integer (i.e., x=0, x=1, etc.). Suppose f is reflection across the vertical line at x = 0, and suppose g slides the strip to the right 2 units. If we do g then f, the H at x=0 first goes to x=2, then goes to x=-2. If we do f then g, the H at x=0 at first stays at x=0, then goes to x=2. Thus gf is not equal to fg.

- [4] Use Euclid's algorithm (described in Chapter 0) to find a solution
(m, n) of 13m + 8n = 1.
13 = 1(8) + 5 so 5 = 13 - 8 8 = 1(5) + 3 so 3 = 8 - 5 = 8 - (13 - 8) = 2(8) - 13 5 = 1(3) + 2 so 2 = 5 - 3 = 13 - 8 - (2(8) - 13) = 2(13) - 3(8) 3 = 1(2) + 1 so 1 = 3 - 2 = 2(8) - 13 - (2(13) - 3(8)) = -3(13) + 5(8) 2 = 2(1) + 0 Thus we see that 1 = gcd(13,8) = -3(13) + 5(8), so we can take m = -3 and n = 5. There are infinitely many other solutions: (m,n) = (-3, 5), (-3 + 8, 5 - 13), (-3 + 2(8), 5 - 2(13)), (-3 + 3(8), 5 - 3(13)), etc., all work.