Let be a nontrivial finite abelian group of rank .
- Prove that the rank of is the maximum of the ranks of its Sylow subgroups.
- Prove that can be generated by elements, but that no subset of fewer than elements generates .
- Suppose (via FTFGAG) has type , where for . Then without loss of generality, . By a previous exercise, every Sylow -subgroup of is of the form , where is a Sylow -subgroup. Again by FTFGAG, the type of is , where is maximal such that . In particular, the rank of is at most . Moreover, if divides , then and the rank of is . Thus the rank of is the maximum of the ranks of is Sylow subgroups.
- We begin with a lemma.
Lemma: The elementary abelian group of order is not generated by any subset of fewer than elements. Proof: Let be a subset consisting of elements. Now because is commutative and every element has order , every element of can be written in the form , where . Thus there are at most elements in , a contradiction.
Again without loss of generality, , where . Denote by the element whose th component is and whose other components are 1. Clearly . Thus is generated by elements.
Suppose now that is generated by a subset containing elements. By part (1), there is a prime such that the (necessarily unique) Sylow -subgroup of has rank . Let denote the -power map on ; by a previous exercise, . But then is generated by elements, a contradiction of the lemma.