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Finite abelian group into sequence of subgroups

by jav
Tags: abelian, finite, sequence, subgroups
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jav
#1
Mar2-12, 01:28 AM
P: 35
G finite abelian group

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

My original thought was to create Hi+1 by reducing the power of one of the generators of Hi by a prime p. Then the order of Hi/Hi+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.

(kiHi+1)(hiHi+1)(kiHi+1)-1 = (hiHi+1) where hi c Hi, ki c Ki c Hi

since G abelian implies the inverse coset commutes.

Then, in order to prove Hi/Hi+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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Deveno
#2
Mar2-12, 02:02 PM
Sci Advisor
P: 906
there are a couple of "easier" cases you might want to look into first:

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

|G| = pk, 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?).


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