Let G be a group, , where p is prime.

Prove G has a sub-group for each of the order , where

Can someone provide me hints/sketches to attempt this? Have been stuck for a long time now.

Thanks

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- Aug 15th 2009, 05:08 AMaman_ccGroup Theory - Basic question on existence of sub-groups
Let G be a group, , where p is prime.

Prove G has a sub-group for each of the order , where

Can someone provide me hints/sketches to attempt this? Have been stuck for a long time now.

Thanks - Aug 15th 2009, 06:39 AMynj
use induction on n

for any k<n, if p^k<=|Z(G)|, then Z(G) have a subgroup of order p^k since it is abelian. if p^k>|Z(G)|,then by induction,G/Z(G) must have a subgroup of order p^k/|Z(G)|,by the corresponding theorem, we can write it as H/Z(G), then H will be a subgroup of order p^k - Aug 15th 2009, 07:24 AMaman_cc
- Aug 15th 2009, 08:42 AMynj
There is the theorem: |G|=n ,G is an abelian, m|n, then G has a subgroup of order m.

Proof:

G is isomorphic to H=Z(p1^k1)*...Z(pt^kt), where pi is prime.

let n=p1^k1*p2^k2...*pt^kt,m=p1^l1....*pt^lt, where li<=ki; f be the isomorphism of H to G

then the group generated by element f(p1^(k1-l1),0,0..0),f(0,p2^(k2-l2),0...0).....f(0..0,pt^(kt-lt))will have order p1^l1*...pt^lt=m - Aug 15th 2009, 08:52 AMynj
Actually, you can assert that for every p^k||G|,where p is a prime, then G has a subgroup of order p^k according to the First Sylow Theorem

- Aug 15th 2009, 04:39 PMNonCommAlg
proof by induction over : there's nothing to prove if (or ). now consider two cases:

1) is abelian: by Cauchy's theorem has an element of oreder now apply induction to the abelian group

2) General case: let where if we're done by 1). if let we have so, by induction, has a subgroup

of order which gives us done again! - Aug 16th 2009, 12:36 AMaman_cc