Prime order implies no proper nontrivial subgroup

Statement
If a finite group has order $$p$$ for some prime number $$p$$, then it has no proper nontrivial subgroup. In other words, the only possible subgroups of the group are the trivial subgroup and the whole group.

Facts used

 * 1) uses::Lagrange's theorem: In the simplistic form here, the order of any subgroup of a group divides the order of the group.

Proof
Given: A group $$G$$ of order equal to a prime number $$p$$. A subgroup $$H$$ of $$G$$.

To prove: $$H$$ is either equal to $$G$$ or the trivial subgroup.

Proof: By fact (1), the order of $$H$$ divides the order of $$G$$, which is the prime number $$p$$. Thus, $$H$$ has order equal to $$1$$ or $$p$$. We consider both cases:


 * If $$H$$ has order $$1$$, it must be the trivial subgroup, because $$H$$ must contain the identity element.
 * If $$H$$ has order $$p$$, it has to equal the whole group $$G$$.