
#1
Mar212, 01:28 AM

P: 35

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? 



#2
Mar212, 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 = 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?). 


Register to reply 
Related Discussions  
Subgroups of a finitely generated abelian group  Calculus & Beyond Homework  4  
Group action on cosets of subgroups in nonabelian groups  Linear & Abstract Algebra  2  
prove that a group of order (m^2)(n^2) is abelian if it has all normal subgroups  Calculus & Beyond Homework  14  
Abelian Group/Subgroups of Power Algebra  Calculus & Beyond Homework  5  
Cyclic subgroups of an Abelian group  Calculus & Beyond Homework  1 