Let be a group, and let be an element of finite order; say . Use the Division Algorithm to show that any integral power of is equal to one of the elements in the set . Conclude that is precisely the set of distinct elements of the cyclic subgroup of generated by .
Let . By the Division Algorithm, there exist unique integers such that and . Thus , where .
Hence the cyclic subgroup of generated by is ; moreover, by a previous result, the elements of are all distinct.
Let . By the Division Algorithm, there exist unique integers such that and . Thus , where .
Hence the cyclic subgroup of generated by is ; moreover, by a previous result, the elements of are all distinct.
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