Finite group with identity subgroup.

Do you agree that, by definition, a group $G$ is cyclic iff there is some $a$ in $G$ such that $<a>=G$ .

Quote:

I know that the order of this group is 1

Since the order of a finite group is its cardinality, the one and only group of order 1 is the trivial group $G=\{1_G\}.$ But there are non trivial groups with no subgroup other than $\{1\}$ and themselves (i.e. with no non-trivial subgroup).

So, consider a finite group $G\neq\{1_G\}$ with no non-trivial subgroup, let's say $G$ has order $n\geq 2$ . You may know that if a prime $p$ divides $n,$ then $G$ has an element of order $p.$
Assume $n$ is not a prime, and try to obtain a contradiction by proving $G$ 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 $G=\{1_G\}=<1_G>$ therefore $G$ is cyclic, because its only element is a generator. If it is not 1, take an element of order strictly greater than $1$ (justify its existence) and consider the subgroup it spans. Because of the hypothesis, since it is different from $\{1_G\},$ it must be the whole group, hence $G$ 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.