Theorem·T16
Groups of prime order are cyclic
A group whose size is a prime is generated by any one of its non-identity elements, and has no proper nontrivial subgroups.
Let
be a group whose order
(cardinality) is prime. Then
is cyclic:
Consequently
has no subgroups other than
and
.
In words
For every g, if g is an element of G other than the identity then the cyclic subgroup generated by g is all of G: G has no subgroups other than the trivial one and itself.
Never needed: F10 · A03 · A04 · A05 · A07 (computed from the citation graph, not asserted).
Proof
- 1
- 2The cyclic subgroup has (L35), and Lagrange's theorem gives . As is prime its only divisors are and , so .
- 3If then , and (L32 (iii)), forcing , contrary to . Hence .
- 4So with ; L34 gives . Thus every element of is a power of : is cyclic, generated by .
- 5
∎
Remarks
The simplest nontrivial groups. Having no room for a proper nontrivial subgroup (a corollary of Lagrange) forces such a group to be cyclic. Any two groups of the same prime order are isomorphic, to the additive group of integers modulo
, once the integers and modular arithmetic are built; this is the base case of the classification of finite abelian groups, and shows there is exactly one group of each prime order.