- #1
jav
- 35
- 0
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?
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?
Last edited: