finite subgroup
Proof. The condition (1) is apparently true if is a subgroup. Conversely, suppose that a nonempty finite subset of the group satisfies (1). Let and be arbitrary elements of . By (1), all () powers of belong to . Because of the finiteness of , there exist positive integers such that
By (1),
Thus also , whence, by the theorem of the http://planetmath.org/node/1045parent entry, is a subgroup of .
Example. The multiplicative group![]()
of all nonzero complex numbers
![]()
has the finite multiplicative subset
, which has to be a subgroup of .
| Title | finite subgroup |
|---|---|
| Canonical name | FiniteSubgroup |
| Date of creation | 2013-03-22 18:57:02 |
| Last modified on | 2013-03-22 18:57:02 |
| Owner | pahio (2872) |
| Last modified by | pahio (2872) |
| Numerical id | 5 |
| Author | pahio (2872) |
| Entry type | Theorem |
| Classification | msc 20A05 |
| Synonym | criterion for finite subgroup |
| Synonym | finite subgroup criterion |