G finite abelian group
WTS: There exist sequence of subgroups {e} = H_{r} c .... c H_{1} 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_{i}H_{i+1})(h_{i}H_{i+1})(k_{i}H_{i+1})^{1} = (h_{i}H_{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?
