# No proper nontrivial subgroup implies cyclic of prime order

## Statement

If a nontrivial group has no proper nontrivial subgroup, then it is a cyclic group of prime order. In other words, it is generated by a single element whose order is a prime number.

## Converse

`Further information: Prime order implies no proper nontrivial subgroup`

The converse is clearly true from Lagrange's theorem. Namely, if we have a cyclic group of prime order , any subgroup must have order either 1 or , by Lagrange's theorem. The only possible subgroup of order is the whole group, and the only possible subgroup of order 1 is the trivial subgroup.

Note that this also shows something stronger: *any* group of prime order, since it has no proper nontrivial subgroup, must be cyclic of prime order.

## Related facts

### Corollaries

- Cyclic of prime power order iff not generated by proper subgroups
- Every subgroup is a direct factor iff trivial or elementary Abelian
- Cyclic iff not a union of proper subgroups

## Proof

**Given**: A nontrivial group , such that the only subgroups of are the trivial subgroup, and itself

**To prove**: There exists an element such that , and the order of is a prime number. In particular, with

**Proof**: Since is nontrivial, there exists in . Then, consider the subgroup generated by . This is a nontrivial subgroup, hence, by assumption, .

Now there are two possibilities. First, that has infinite order: no positive power of is trivial. In this case, the group is isomorphic to (i.e., can be identified with) the group , under the identification . In particular, the subgroup generated by , which corresponds to the even integers, is a proper nontrivial subgroup, leading to a contradiction.

Hence, cannot have infinite order, so let be the order of . Then, . Suppose that is composite. Then, is isomorphic to the cyclic group of order , under the identification . Pick a positive integer such that . Then the subgroup generated by is a proper nontrivial subgroup of (corresponds to the multiples of mod ). More precisely, it is a subgroup of order , because the order of is . This is again a contradiction.

Hence, must be a prime, so is cyclic of prime order (as desired).