The rank of a finite abelian group is the maximum rank of its Sylow subgroups

Let G be a nontrivial finite abelian group of rank t.

  1. Prove that the rank of G is the maximum of the ranks of its Sylow subgroups.
  2. Prove that G can be generated by t elements, but that no subset of fewer than t elements generates G.

  1. Suppose (via FTFGAG) G has type (n_1,n_2,\ldots,n_t), where n_i|n_j for i \geq j. Then without loss of generality, G = Z_{n_1} \times \cdots \times Z_{n_t}. By a previous exercise, every Sylow p-subgroup of G is of the form P = P_1 \times \cdots \times P_t, where P_i \leq Z_{n_i} is a Sylow p-subgroup. Again by FTFGAG, the type of P is (|P_1|, \ldots, |P_s|), where s is maximal such that P_s \neq 1. In particular, the rank of P is at most t. Moreover, if p divides n_t, then s = t and the rank of P is t. Thus the rank of G is the maximum of the ranks of is Sylow subgroups.
  2. We begin with a lemma.

    Lemma: The elementary abelian group E of order p^t is not generated by any subset of fewer than t elements. Proof: Let S \subseteq E be a subset consisting of k < t elements. Now because E is commutative and every element has order p, every element of E can be written in the form s_1^{a_1} s_2^{a_2} \cdots s_k^{a_k}, where 0 \leq a_i < p. Thus there are at most p^k < p^t elements in E, a contradiction. \square

    Again without loss of generality, G = Z_{n_1} \times \cdots \times Z_{n_t}, where Z_{n_i} = \langle x_i \rangle. Denote by e_i \in G the element whose ith component is x_i and whose other components are 1. Clearly G = \langle e_1, e_2, \ldots, e_t \rangle. Thus G is generated by t elements.

    Suppose now that G is generated by a subset S containing k < t elements. By part (1), there is a prime p such that the (necessarily unique) Sylow p-subgroup P of G has rank t. Let \varphi denote the p-power map on G; by a previous exercise, G/\mathsf{im}\ \varphi \cong Z_p^t. But then Z_p^t is generated by k < t elements, a contradiction of the lemma.

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Comments

  • will  On November 3, 2011 at 6:59 pm

    Hi,

    Thanks for the incredibly useful site!

    A question about part 1:
    Why must there necessarily be a Sylow p-subgroup such that p divides n_t? That is, why can it not be the case that the rank of the group is greater than that of any of its Sylow p-subroups?

    • will  On November 3, 2011 at 7:11 pm

      Just answered my own question—

      It follows because there is at least one Sylow p-subgroups for every prime p, and the last invariant factor of G, n_t, is either a prime or a product of primes.

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