2.2.20

a) Because the group is Abelian $$(ab)^n = a^n b^n$$. To see this consider the following examples: $$(ab)^2 = abab = a^2b^2$$, $$(ab)^3 = ababab = a^3b^b$$ and so on (i.e., inductively).

The element $$ab$$ will have finite order. To see this, consider $$n = 2, m = 3$$ and let $$\{ a_0, a_1 \}$$ denote the group generated by $$a$$, and $$\{ b_0, b_1, b_2 \}$$ denote the group generated by $$b$$. Using this notation the group generated by $$ab$$ is $$ = \{a_0b_0, a_1b_1, a_0b_2, a_1b_0, a_0b_1, a_1b_2 \}$$. Because some elements in $$$$ might be equal, we cannot say the exact order of the group. However, because the order of $$ab$$ is bounded by $$nm$$ (in general), we can at least say that $$ab$$ is of finite order.