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 $\displaystyle x\in O_p(G)$: 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. $\displaystyle H\slash O_p(G)\le G\slash O_p(G)$[/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) $\displaystyle H\slash O_p(G)$ is a Sylow p-sbgp. then..." , and $\displaystyle O_p(G)\le H$ 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 $\displaystyle N\triangleleft G$ and $\displaystyle P$ is a Sylow p-sbgp of G, then $\displaystyle PN\slash N$ is a Sylow p-sbgp. of $\displaystyle G\slash N$, so $\displaystyle H\slash O_p(G)$ a Syl. p-sbgp of $\displaystyle G\slash O_p(G)\Longrightarrow H=PO_p(G)$ , with $\displaystyle P$ some Syl. p-sbgp. of $\displaystyle G$...but $\displaystyle O_p(G)\le P$ for ANY Syl. p-sbgp $\displaystyle P$ of $\displaystyle G\Longrightarrow H=PO_p(G)= P$ 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!