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

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

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

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

Thanks

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

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

3. 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

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

4. 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

5. 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

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

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

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

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

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

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

7. Thanks. I get the idea.

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

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

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

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