A question from GRE math sub 9367, problem 59?

Question

Two subgroups H and K of a group G have orders 12 and 30, respectively. Which of the following could NOT be the order of the subgroup of G generated by H and K? A. 30 B. 60 C. 120 D. 360 E. Countable infinity A is the answer because H, with order 12 that doesn't divide 30, can't be a subgroup of K. But anybody can help me construct a concrete example of E, the subgroup generated by H and K with order of countable infinity?

Answer 1

Let G be the group defined by generators a,b, and relations a^{12}=b^{30}=1. Then H=<a> and K=<b> are two subgroups of G of orders 12 and 30, respectively. However, the subgroup of G generated by H and K is G itself, whose order is countably infinite: a and b do not commute, and they can generate elements of G given by non-reducible words of arbitrary finite length. It is then an easy exercise to check the desired conclusion.