Abstract Algebra (3th ed.)/Chapter1/Exercises 1.2

05.03.2020

1)

In general $|s| = 2$ by (2). And we can compute

1. For $D_6$ we have the following elements $$\{ 1,r,r^2 , s , sr , sr^2 \}$$ Now $|1| = 1, |r| = 3$. And we have that $$(r^2)^2 = r^4 = r$$ So $|r^2| = 3$. $s(rs)r = ssr^{-1}r = 1$, so $|sr| = 2$. We have that $$s (r^2 s) r^2 = ss r^{-2}r^2 = 1$$ So $|sr^2| = 2$. In total
 1 r r^{2} s sr sr^{2} 1 3 3 2 2 2
2. For $D_8$ we have the following elements $$\{ 1,r,r^2,r^3,s,sr,sr^2,sr^3 \}$$ here $|r| = 4$. We have that $(r^2)^2 = 1$. So $|r^2| = 2$. We have that $|r^3| = 4$. We have that every element on the form $sr^i$ has order $2$. That is $$s(r^i s) r^i = ss r^{-i} r^{i} = 1$$ So in total
 1 r r^{2} r^{3} s sr sr^{2} sr^{3} 1 4 2 4 2 2 2 2
3. For $D_{10}$ we have the following elements $$\{ 1,r,r^2,r^3,r^4,s,sr,sr^2,sr^3,sr^4 \}$$ We have $|r| = 5$. We have that $|r^2| = 5$. We have that $|r^3| = 5$. And $|r^4| = 5$. In total we get
 1 r r^{2} r^{3} r^{4} s sr sr^{2} sr^{3} sr^{4} 1 5 5 5 5 2 2 2 2 2