From here it should be clear why (a) and (b) follows.
But you can also see (a) because if then it means that for all . But if then if or if . Therefore and so the complex number in must lie on the unit circle. To show (b) let . By Lagrange's theorem we have . Therefore, for some . Define , this subgroup is cyclic and contains , but, , from here it follows that . Thus, must be a cyclic subgroup.