# Thread: Abstract-Sylow THEOREM

1. ## Abstract-Sylow THEOREM

Hey there,

Help is needed in the next question:

Let G be a finite group of order n and let p be a prime number that divides n.
Let's mark as Op(G) as the intersection of all p-sylow groups of G.

1. Prove that Op(G) is a normal p-subgroup in G.
2. Prove that every normal p-subgroup of G is contained in Op(G)
3. Let's mark: G] = G/Op(G). Prove that Op(G] ) ={1}

About 1- It's easy to see that it's a subgroup...I'm not sure about what to do in the normality proof... I need help in all the other problems too....

TNX everyone!

2. Originally Posted by WannaBe
Hey there,

Help is needed in the next question:

Let G be a finite group of order n and let p be a prime number that divides n.
Let's mark as Op(G) as the intersection of all p-sylow groups of G.

1. Prove that Op(G) is a normal p-subgroup in G.
2. Prove that every normal p-subgroup of G is contained in Op(G)
3. Let's mark: G] = G/Op(G). Prove that Op(G] ) ={1}

About 1- It's easy to see that it's a subgroup...I'm not sure about what to do in the normality proof... I need help in all the other problems too....

TNX everyone!

If P is a Sylow p-sbgp. of G, then ALL the other Sylow p-sbgps. are of the form $g^{-1}Pg\,,\,g\in G$ , and from here you get normality of $O_p(G)$ at once.

If H is a normal p-sbgp. of G then H is contained in each end every Sylow p-sbgp of G (why? Read above)

If $H\slash O_p(G)$ is a Syl p-sbgp. of $G\slash O_P(G)$ then:

1) $O_p(G)\le H$

2) $\pi^{-1}(H)\le G$ , with $\pi:G\rightarrow G\slash O_p(G)$ the canonical projection, is a p-sbgp. of G...but we know that EVERY p-sbgp. of G is contained in some Sylow p-sbgp. of G...

Tonio

3. Something is still missing:
1. By sylow's 3rd theorem, we know that each 2 p-sylow subgroups are conjugates...To prove that the sub-group Op(G) is normal, we need to prove, that for each g in G we get: g^-1 * Op(G) * g = Op(G)... And it isn't what you've written...So we still have to prove that it's a normal sbgrp. ... We also need to prove that it's a p group...!
About 2 - Completely understood...
I didn't completely understand 3... Let H be a normal p-sbrgp. of G... Why you've said that H/Op(G) is a Syl p-sbgp. of G/Op(G)??Why you've said afterwards that Op(G)<H?
On which group you've said that it's a p-sbgp of G??
As you can see, I didn't quite understand 3 and 1... I will be delighted if you'll be able to help me...

TNX!

4. Originally Posted by WannaBe
Something is still missing:
1. By sylow's 3rd theorem, we know that each 2 p-sylow subgroups are conjugates...To prove that the sub-group Op(G) is normal, we need to prove, that for each g in G we get: g^-1 * Op(G) * g = Op(G)... And it isn't what you've written...

Of course that's not what I wrote: I gave you some help, now you complete the task...

So we still have to prove that it's a normal sbgrp

That follows from the above at once: think.

. ... We also need to prove that it's a p group...!

...common! Take an element $x\in O_p(G)$: what MUST be its order?

About 2 - Completely understood...
I didn't completely understand 3... Let H be a normal p-sbrgp. of G...

Why normal?? The normal H I wrote about was part of question 2...!
For three you only need a quotient sbgp. $H\slash O_p(G)\le G\slash O_p(G)$[/color]

Why you've said that H/Op(G) is a Syl p-sbgp. of G/Op(G)??Why you've said afterwards that Op(G)<H?

I didn't say such a thing: I wrote: "IF (IIIFFF) $H\slash O_p(G)$ is a Sylow p-sbgp. then..." , and $O_p(G)\le H$ because this much is true in any quotient group.
Perhaps it'd be a good idea you take a peek again on this stuff: remember for example that if $N\triangleleft G$ and $P$ is a Sylow p-sbgp of G, then $PN\slash N$ is a Sylow p-sbgp. of $G\slash N$, so $H\slash O_p(G)$ a Syl. p-sbgp of $G\slash O_p(G)\Longrightarrow H=PO_p(G)$ , with $P$ some Syl. p-sbgp. of $G$...but $O_p(G)\le P$ for ANY Syl. p-sbgp $P$ of $G\Longrightarrow H=PO_p(G)= P$ is a Syl. p-sbgp. of G...
Perhaps reading a little about subgroups of quotient groups (e.g., correspondence theorem) would help, too.

Tonio

On which group you've said that it's a p-sbgp of G??
As you can see, I didn't quite understand 3 and 1... I will be delighted if you'll be able to help me...

TNX!
.

5. Well let's see:
1. Let P be a SylSbg. and let k be an element in Op(G)...
that means that there exists g in G such as for an element p in P : k=g^-1*p*g...
We need to prove that for each f in G : f^-1 * k* f is also in Op(G) . But:
f^-1*k*f=(gf)^-1 *k * (gf) which is also an element of all the P syl.Sbg....So we've proved it's normal...
About the order thing: each element in Op(G) has an order p^i for some i in N...Is this tells us that it's a p-group?

About the rest- Completelty understandable...
TNX a lot! I'll be delighted if you'll be able to verify my answer ...

6. Originally Posted by WannaBe
Well let's see:
1. Let P be a SylSbg. and let k be an element in Op(G)...
that means that there exists g in G such as for an element p in P : k=g^-1*p*g...
We need to prove that for each f in G : f^-1 * k* f is also in Op(G) . But:
f^-1*k*f=(gf)^-1 *k * (gf) which is also an element of all the P syl.Sbg....So we've proved it's normal...
About the order thing: each element in Op(G) has an order p^i for some i in N...Is this tells us that it's a p-group?

Of course it does: it is the definition of p-group! In case the group is finite, this is equivalent with saying that the order of the group is a power of p...
And your solution for normality of $O_p(G)$ is correct.

Tonio

About the rest- Completelty understandable...
TNX a lot! I'll be delighted if you'll be able to verify my answer ...
.