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Saturday, March 16, 2013

Finite Abelian Groups and Rank (5.2.11)

Dummit and Foote Abstract Algebra, section 5.2, exercise 11: MathJax TeX Test Page

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 PSylp(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 φker φ. Prove that ker φ=ker φ|P. () If xker φ|P, then definitionally xp=1 and xPG. () Assume xp=1 but xP. 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 xP. Therefore ker φ=ker φ|P. By part (a), we have that ker φ|PEpt, so that G/im φEpt, 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 KEpt 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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