No proper nontrivial subgroup implies cyclic of prime order

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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 p, any subgroup must have order either 1 or p, by Lagrange's theorem. The only possible subgroup of order p 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


Other related facts

Proof

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

To prove: There exists an element g \in G such that \langle g \rangle = G, and the order of g is a prime number. In particular, G = \{ e, g, g^2, \dots, g^{p-1} \} with g^p = e

Proof: Since G is nontrivial, there exists g \ne e in G. Then, consider the subgroup generated by g. This is a nontrivial subgroup, hence, by assumption, \langle g \rangle = G.

Now there are two possibilities. First, that g has infinite order: no positive power of g is trivial. In this case, the group G is isomorphic to (i.e., can be identified with) the group \mathbb{Z}, under the identification n \mapsto g^n. In particular, the subgroup generated by g^2, which corresponds to the even integers, is a proper nontrivial subgroup, leading to a contradiction.

Hence, g cannot have infinite order, so let n be the order of g. Then, g^n = e. Suppose that n is composite. Then, G is isomorphic to the cyclic group of order n, under the identification a \mapsto g^a. Pick a positive integer d \ne 1,n such that d | n. Then the subgroup generated by g^d is a proper nontrivial subgroup of G (corresponds to the multiples of d mod n). More precisely, it is a subgroup of order n/d, because the order of g^d is n/d. This is again a contradiction.

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