# Group Theory - Basic question on existence of sub-groups

• Aug 15th 2009, 05:08 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
• Aug 15th 2009, 06:39 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
• Aug 15th 2009, 07:24 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

• Aug 15th 2009, 08:42 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
• Aug 15th 2009, 08:52 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
• Aug 15th 2009, 04:39 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 $\frac{G}{}.$

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!
• Aug 16th 2009, 12:36 AM
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 $\frac{G}{}.$

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!