suppose so that . then as you can verify by multiplication.
therefore, , so is finite.
so if .
now suppose that , so that .
(since G is abelian)
, so , that is .
note we didn't show what the order of actually was, just that it was finite (and less than or equal to mn).
finding a counter-example is a bit tricky. let's use G = Sym(Z), the group of all bijections of Z.
define f(n) = -n. it's clear f is a bijection, and f has order 2.
define g(n) = 1-n. it's also clear g is a bijection, and:
g(g(n)) = g(1-n) = 1 - (1-n) = 1 - 1 + n = n, so g also has order 2.
but fg(n) = f(g(n)) = f(1-n) = n-1, and for any positive integer k:
so fg is of infinite order.