Free abelian group

Subgroup of a free abelian group

Suppose is a free abelian group of rank and . Then: #m/thm/group

1. is free of rank ;
2. There exists a basis for and integers such that forms a basis for ;
3. The Lagrange index is finite iff .

Moreover, if we have a change of basis from a basis of to of such that

Then .¹

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