Group Theory - Basic question on existence of sub-groups

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August 15th 2009, 04:08 AM
aman_cc

Group Theory - Basic question on existence of sub-groups

Let G be a group, $\mid G \mid = p^n$ , where p is prime.
Prove G has a sub-group for each of the order $p^\alpha$ , where $\alpha \in [0,n]$

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

Thanks
August 15th 2009, 05:39 AM
ynj

Quote:

Originally Posted by aman_cc

Let G be a group, $\mid G \mid = p^n$ , where p is prime.
Prove G has a sub-group for each of the order $p^\alpha$ , where $\alpha \in [0,n]$

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
August 15th 2009, 06:24 AM
aman_cc

Sorry but a little elaboration will help, not able to follow the argument. Maybe you can mention the relevant theorm for e.g. I am not able to follow why "if p^k<=|Z(G)|, then Z(G) have a subgroup of order p^k since it is abelian"
Thanks

Quote:

Originally Posted by ynj

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
August 15th 2009, 07:42 AM
ynj

Quote:

Originally Posted by aman_cc

Sorry but a little elaboration will help, not able to follow the argument. Maybe you can mention the relevant theorm for e.g. I am not able to follow why "if p^k<=|Z(G)|, then Z(G) have a subgroup of order p^k since it is abelian"
Thanks

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
August 15th 2009, 07:52 AM
ynj

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
August 15th 2009, 03:39 PM
NonCommAlg

Quote:

Originally Posted by aman_cc

Let G be a group, $\mid G \mid = p^n$ , where p is prime.
Prove G has a sub-group for each of the order $p^\alpha$ , where $\alpha \in [0,n]$

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

Thanks

proof by induction over $n$ : there's nothing to prove if $n=0$ (or $\alpha = n$ ). now consider two cases:

1) $G$ is abelian: by Cauchy's theorem $G$ has an element $x$ of oreder $p.$ now apply induction to the abelian $p$ group $\frac{G}{<x>}.$

2) General case: let $|Z(G)|=p^m,$ where $0 < m \leq n.$ if $\alpha \leq m,$ we're done by 1). if $\alpha > m,$ let $G_1=\frac{G}{Z(G)}.$ we have $|G_1|=p^{n-m} < p^n.$ so, by induction, $G_1$ has a subgroup $H_1=\frac{H}{Z(G)}$

of order $p^{\alpha-m},$ which gives us $|H|=p^{\alpha}.$ done again!
August 15th 2009, 11:36 PM
aman_cc

Thanks. I get the idea.

Quote:

Originally Posted by NonCommAlg

proof by induction over $n$ : there's nothing to prove if $n=0$ (or $\alpha = n$ ). now consider two cases:

1) $G$ is abelian: by Cauchy's theorem $G$ has an element $x$ of oreder $p.$ now apply induction to the abelian $p$ group $\frac{G}{<x>}.$

2) General case: let $|Z(G)|=p^m,$ where $0 < m \leq n.$ if $\alpha \leq m,$ we're done by 1). if $\alpha > m,$ let $G_1=\frac{G}{Z(G)}.$ we have $|G_1|=p^{n-m} < p^n.$ so, by induction, $G_1$ has a subgroup $H_1=\frac{H}{Z(G)}$

of order $p^{\alpha-m},$ which gives us $|H|=p^{\alpha}.$ done again!