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Math Help - Abstract-Sylow THEOREM

  1. #1
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    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!
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  2. #2
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    Quote Originally Posted by WannaBe View Post
    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
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  3. #3
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    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!
    Last edited by WannaBe; December 11th 2009 at 07:00 AM.
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  4. #4
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    Quote Originally Posted by WannaBe View Post
    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!
    .
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  5. #5
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    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 ...
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  6. #6
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    Quote Originally Posted by WannaBe View Post
    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 ...
    .
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