Let G be a finite abelian group of rank t.
(a) Prove that the rank of G is the maximum of the ranks of G's Sylow subgroups.
(b) Prove that the smallest subset of G generating G is of order t. [One way of doing this is by combining part (a) with exercise 7.]
Proof: (a) Obtain G's invariant factor decomposition, and let p be a prime dividing ns, the order of the smallest cyclic factor. By 5.1.4, we have the invariant factor decomposition of P∈Sylp(G) is of rank t.
(b) We know G is generated by the t generators of each of the cyclic factors of G, so existence is not an issue. Assume that G=<S>, where |S|=s<t.
Obtain P of rank t by part (a), and let φ be the p-power map. By 5.2.7(a), we have G/im φ≅ker φ. Prove that ker φ=ker φ|P. (⊇) If x∈ker φ|P, then definitionally xp=1 and x∈P≤G. (⊆) Assume xp=1 but x∉P. Since every p-subgroup is contained within a Sylow p-subgroup, and P is the unique Sylow p-subgroup of G by the abelian quality of G, we have <x>≤P, and thus x∈P. Therefore ker φ=ker φ|P. By part (a), we have that ker φ|P≅Ept, so that G/im φ≅Ept, and in particular - as a consequence of the former being generated by the natural projection image of S - the latter is generatable by a subset of fewer than t elements.
For a generating subset K⊆Ept and |K|=n<t, we have every element of Ept is rewritable as kα11kα22...kαnn. Since every nontrivial element of the elementary abelian group is of order p, we have an upper bound of pn distinct elements writable in this fashion, a contradiction since pn<pt=|Ept|. ◻
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