G finite abelian group(adsbygoogle = window.adsbygoogle || []).push({});

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?

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# Finite abelian group into sequence of subgroups

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