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 |