Do you agree that, by definition, a group is cyclic iff there is some in such that .

Since the order of a finite group is its cardinality, the one and only group of order 1 is the trivial group But there are non trivial groups with no subgroup other than and themselves (i.e. with no non-trivial subgroup).I know that the order of this group is 1

So, consider a finite group with no non-trivial subgroup, let's say has order . You may know that if a prime divides then has an element of order

Assume is not a prime, and try to obtain a contradiction by proving has a non-trivial subgroup.

So far you'll have that a finite group with no non-trivial subgroup has order 1 or a prime. If it is 1, do you agree that therefore is cyclic, because its only element is a generator. If it is not 1, take an element of order strictly greater than (justify its existence) and consider the subgroup it spans. Because of the hypothesis, since it is different from it must be the whole group, hence is cyclic.

Note that we did not to use that order (when different from 1) is prime; that being said a similar argument shows that any group with order a prime has no non-trivial subgroup (and is cyclic).

Conclusion: A group has no non-trivial subgroup iff its order is 1 or a prime.