Solution: To show that a group \( G \) with no proper nontrivial subgroups is cyclic, we begin by considering the nature of such a group. If \( G \) is not the trivial group, then it must contain at least one element \( g \) not equal to the identity element.

Since \( G \) has no proper nontrivial subgroups, the subgroup generated by \( g \), denoted by \( \langle g \rangle \), must be either \( G \) itself or the trivial subgroup. However, since \( g \) is not the identity element, \( \langle g \rangle \) cannot be trivial, implying that \( \langle g \rangle = G \).

Therefore, every element of \( G \) can be expressed as a power of \( g \), proves that \( G \) is cyclic. Hence, we conclude that if a group has no proper nontrivial subgroups, it must be cyclic.