Finite abelian group into sequence of subgroups

jav

G finite abelian group

WTS: There exist sequence of subgroups {e} = H_r c .... c H₁ c G
such that H_i/H_i+1 is cyclic of prime order for all i.

My original thought was to create H_i+1 by reducing the power of one of the generators of H_i by a prime p. Then the order of H_i/H_i+1 would be p, but not necessarily cyclic.

I also know that a simple finite abelian group is cyclic of prime order, but don't know how to construct simple cosets.

(k_iH_i+1)(h_iH_i+1)(k_iH_i+1)^-1 = (h_iH_i+1) where h_i c H_i, k_i c K_i c H_i

since G abelian implies the inverse coset commutes.

Then, in order to prove H_i/H_i+1 is simple would be equivalent to showing that the trivial subgroup and itself are the only subgroups. If that were true, then there would only be one valid subgroup of G in the sequence. Ie. the sequence would look like {e} c H c G.

What am I missing here?

Deveno

there are a couple of "easier" cases you might want to look into first:

|G| = pq, p,q distinct primes.

|G| = p^k, p a prime.

the second case is "harder", although you may have proved both of these already if you have covered the sylow theorems.

by the way, any group of prime order is necessarily cyclic (why?).