Editing Subgroup (section)

==Subgroup tests==

Suppose that {{mvar|G}} is a group, and {{mvar|H}} is a subset of {{mvar|G}}.  For now, assume that the group operation of {{mvar|G}} is written multiplicatively, denoted by juxtaposition.
*Then {{mvar|H}} is a subgroup of {{mvar|G}} [[if and only if]] {{mvar|H}} is nonempty and [[Closure (mathematics)|closed]] under products and inverses. ''Closed under products'' means that for every {{mvar|a}} and {{mvar|b}} in {{mvar|H}}, the product {{mvar|ab}} is in {{mvar|H}}.  ''Closed under inverses'' means that for every {{mvar|a}} in {{mvar|H}}, the inverse {{math|''a''<sup>&minus;1</sup>}} is in {{mvar|H}}. These two conditions can be combined into one, that for every {{mvar|a}} and {{mvar|b}} in {{mvar|H}}, the element {{math|''ab''<sup>&minus;1</sup>}} is in {{mvar|H}}, but it is more natural and usually just as easy to test the two closure conditions separately.{{sfn|Kurzweil|Stellmacher|1998|p=4}}
*When {{mvar|H}} is ''finite'', the test can be simplified: {{mvar|H}} is a subgroup if and only if it is nonempty and closed under products. These conditions alone imply that every element {{mvar|a}} of {{mvar|H}} generates a finite cyclic subgroup of {{mvar|H}}, say of order {{mvar|n}}, and then the inverse of {{mvar|a}} is {{math|''a''<sup>''n''&minus;1</sup>}}.{{sfn|Kurzweil|Stellmacher|1998|p=4}}
If the group operation is instead denoted by addition, then ''closed under products'' should be replaced by ''closed under addition'', which is the condition that for every {{mvar|a}} and {{mvar|b}} in {{mvar|H}}, the sum {{math|''a'' + ''b''}} is in {{mvar|H}}, and ''closed under inverses'' should be edited to say that for every {{mvar|a}} in {{mvar|H}}, the inverse {{math|−''a''}} is in {{mvar|H}}.