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?