Let G be a finite abelian group. I know of two ways of writing it as a direct sum of cyclic groups:

1) With orders d1,d2,…,dk in such a way that di|di+1,

2) With orders that are powers of not necessarily distinct primes pα11,…,pαnn.

Is it true, and how can one prove that the cardinality c of any minimal generating set for G satisfies k≤c≤n (I am most concerned about the second inequality)? Here minimal means irredundant.

**Answer**

Note that n is the sum over prime divisors p of |G| of the minimal number of generators of the distinct Sylow p-subgroups of G. The sizes of all minimal generating sets of a finite p-group are the same by properties of the Frattini subgroup. Use of the Frattini subgroup helps to prove the leftmost inequality: take a prime p which divides d1. Then a Sylow p-subgroup of G can’t be generated by fewer than k elements, so G itself certainly can’t be generated by fewer than k elements, as each Sylow p-subgroup of G is a homomorphic image of G. On the other hand, take a minimal generating set S for G of maximal cardinality, and minimize the sum of the orders of elements of S subject to that. Then each element of S must have prime power order, for if s∈S has order divisible by more than one prime, then we may write s=t+u where t and u have coprime orders (each greater than one) whose product is the order of s. Then (S∖{s})∪{t,u} is still a minimal generating set for G, contradicting the maximality of the cardinality of S. The fact that S is a minimal generating set means that if we now collect the elements of S whose orders are powers of a fixed prime p, we must obtain a generating set for a Sylow p-subgroup of G, and this must be minimal by the choice of S. Hence the cardinality of S is at most n, as defined above.

**Attribution***Source : Link , Question Author : Calc , Answer Author : Geoff Robinson*