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
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
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
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!