# Difference between revisions of "Tour:No proper nontrivial subgroup implies cyclic of prime order"

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## Latest revision as of 19:33, 8 September 2008

This page is part of the Groupprops guided tour for beginners (Jump to beginning of tour)PREVIOUS: Subgroup containment relation equals divisibility relation on generators|UP: Introduction four (beginners)|NEXT: Exploration of cyclic groups

General instructions for the tour | Pedagogical notes for the tour | Pedagogical notes for this part

WHAT YOU NEED TO DO: Understand, thoroughly, the statement and proof of this theorem.

## 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.

- 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).

This page is part of the Groupprops guided tour for beginners. Make notes of any doubts, confusions or comments you have about this page before proceeding.PREVIOUS: Subgroup containment relation equals divisibility relation on generators|UP: Introduction four (beginners)|NEXT: Exploration of cyclic groups