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Theorem 1: Every subgroup of a cyclic group is cyclic. It follows that $$a$$ is of finite order and this in turn implies that $$G$$ is finite, contrary to the hypothesis. Hence $$H$$ must be an infinite cyclic subgroup of $$G$$.Since gcd(32,4) = 4, 32/4 = 8. To find the other elements of order 8, add multiples of 8 to 4: 12, 20, 28. So there are 4 elements of order 8: 4, 12, 20, 28. Each of these elements generate the same cyclic subgroup, so there is only one subgroup of order 8.