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Abstract Alg.-Quotient Groups

  1. Dec 7, 2009 #1
    1. The problem statement, all variables and given/known data
    The problem is:
    Let G be a group of order 12 ( o(G)=12).
    Let's assume that G has a normal sub-group of order 3 and let a be her generator ( <a>=G ).

    In the previous parts of the questions i've proved that:
    1. a has 2 different conjucates in G and o(N(a))=6 or o(N(a))=12 (N(a) is the normalizer of a)
    2. there is a b in G of order 2 who is commutative with a: ab=ba
    3. if o(N(a))=6 then N(a) is a group of order 6 that is generated by ab ( <ab>=N(a) )

    I have no clue in the next 2 parts of the question:
    4. Prove that b is in the center of the group G (C(G)) . There's a clue: notice that b is in
    N(a).
    5. prove that if 2 doesn't divide o(C(G)) then G is isomorphic to A4 (A4 is the group of all even permutations)...

    2. Relevant equations

    None....

    3. The attempt at a solution

    In question 5 , I can't realy understand how the two groups can be isomorphic if A4 has order 13 and G has order 12...
    There's a clue to question 5 too: you can make use of Cayle's theorem...

    I will be delighted if someone will help me...I have no clue in 4 and 5.... detailed answers will be received with happines...
    TNX!
     
  2. jcsd
  3. Dec 7, 2009 #2

    Dick

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    Just to start out, why do you think A4 has 13 elements??
     
  4. Dec 8, 2009 #3
    plz look here:
    http://en.wikipedia.org/wiki/Alternating_group

    at the beginning they wrote about 4th degree alternating group. if I still know how to count, there are 13 elements there...
     
  5. Dec 8, 2009 #4

    Dick

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    A4 = {e, (123), (132), (124), (142), (134), (143), (234), (243), (12)(34), (13)(24), (14)(23)}?? I see 12. Are you counting one of the two cycle pairs twice or something?
     
  6. Dec 8, 2009 #5
    LOL...I didn't count the elelments correctly...Now that we know they both have 12 elements, how can I make homo. using cayley theorem?
    and what about 4?

    TNX
     
  7. Dec 8, 2009 #6

    Dick

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    I hadn't really thought too hard about it yet. It might help if you'd show how you proved 1), 2) and 3). I am a little worried about 5). A4 doesn't have a normal subgroup of order 3. Are you making different assumptions about G for different parts?
     
  8. Dec 8, 2009 #7
    I don't make different assumptions about G in different parts... everything is the same...
    About 1 2 and 3:
    1. a group H is normal if and only if for each g in G and h in H : g^-1*h*g is also in H...
    that means that there are 3 possibilities for conjucates of a...but 1 isn't a conjucate because : g^-1*a*g=1 -> ag=g -> a=1 and this is wrong... so a has at most 2 conjucates... About the order of N(a) : it's easy to prove that if na_ is the number of conjucates of an element a then : na_ = o(G)/o(N(a)) ... we know the order of G, the number of conjucates of a so we know also that o(N(a))=6 or o(N(a))=12...
    2. 2 is a prime number that divides 12 and 6...We know that each group of order X contains sub-groups of a prime order so that the prime order divdes the order of G...
    So there is a subgroup of N(a) from order 2, and it's cyclic of course, so the generator is b...
    3. if a & b commutate then o(ab)=o(a)*o(b)... so we get o(ab)=6 and we also know that ab is in N(a), so ab if from the same order of the sub-group N(a), so it must be cyclic and ab is it's generator....

    did it helped? :( can you help me now?

    TNX again
     
  9. Dec 8, 2009 #8

    Dick

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    For 4, I think you might want to also use that if o(N(a))=6 then N(a) has index 2 in G. That makes it a normal subgroup. I'm still scratching my head over 5. Any ideas?
     
  10. Dec 8, 2009 #9
    Let's if I understood it right...
    b is in N(a) ... if N(a) is from order 12 - then N(a) is G and it's easy to see that it means that b commutate with the rest of G...
    If N(a) if from order 6 - then it's a normal sub-group of G... Which means that for every one of the elements in G-N(a) -> bx=xb (by the definition for normality) ...from here it's easy to see that b commutate with 'a' and with the rest of the elements in G... so b is indeed in C(G)...


    about 5... In the clue that's attached to the question there's that theorem that is an inclusion of Cayley's theorem ( http://en.wikipedia.org/wiki/Cayley's_theorem ) :
    "Let G be group, H a sub-group of G and S the set of right cosets of H in G. In these conditions, there's an homomorphism phi from G to the set of permutations of the elements of S, and the kernel of phi is a normal sub-group in G and the maximal one that is contained in H"...

    I realy have no idea about it...

    Hope you'll be able to help me in the last part of the question and verify my soloution for the 4th part...

    TNX again!
     
  11. Dec 8, 2009 #10

    Dick

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    For 4, yeah, that's it basically. N(a) is normal and b is the only element of N(a) that has order 1. So it must get mapped to itself under conjugation. Still not sure about 5, though.
     
  12. Dec 9, 2009 #11
    Well tnx a lot for your help so far :(
     
  13. Dec 9, 2009 #12
    Wait a second! Now I see that the question is divided into 2 parts: 1 and 2... the first part includes all the 4 first questions, and question 5 is seperated from the 4 previous questions... So, I guess we do make different assumtions...
    The only data we have about 5 is that G is from order 12, and that 2 doesn't divide
    o(C(G)) ... I think it'll be much easier now...
     
  14. Dec 9, 2009 #13
    Well here are some ideas for this part:
    1. 2 doesn't divide o(C(G)). That means that there no normal sub-groups of G of order 3 (If we assume there's a normal sub-group of order 3, we get from the previous parts, that there's a b in C(G) of order 2, and that's wrong...)
    2. from 1, we get that O(C(G))=1... Hence C(G)={1} ... It can help us if we will define an homomorphic function and let her kernel be C(G)...ker(f)={1} -> f is monomorphism (f is injective...)
    3. We know from the theorem I've quoted, that G is homomorphic to the set of permutations of the set S (set of right cosets of a sub-group of G)...

    It's pretty obviuos that we should use Cayley's theorem now, but I realy don't know how to define the homo. funtion so that her kernel will be C(G) and stuff...

    Hope that now you'll be able to push me forward towards the soloution because I'm realy hopeless...

    TNX again !!!
     
  15. Dec 9, 2009 #14

    Dick

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    Well, you are making progress. That's what was confusing me from the beginning. A4 doesn't have a normal subgroup of order 3. It looks like they now want you to take any subgroup of order 3 and form the four cosets S and figure out how G acts on S. Maybe? You can maybe get some more hints by studying A4 (since your G should come out isomorphic to A4). For example, I know A4 can by generated by two elements, one of order 3 and one of order 2. I know it has 8 elements of order 3, and 3 elements of order 2. I know it has 4 nonnormal subgroups of order 3 and 1 normal subgroup of order 4. All of those things must be true for G as well. Gotta run now but I'll try and have another look at this later in the day.
     
  16. Dec 9, 2009 #15
    Hey there....I realy don't think it should be that difficult...I think the hint that tells us to solve using Cayley's theorem will give us the last piece of soloution...But I realy have no idea about the way I should use it...
    Hope you'll be able to help me...

    TNX!
     
  17. Dec 9, 2009 #16

    Dick

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    Ok. Try this. Pick any H to be any subgroup of order 3. There are four right cosets. Elements of G act on those cosets by right multiplication. That gives you a homomorphism of G into S4 (represented as a permutation of the four cosets). Your big job is to show that homomorphism must be injective. Cayley's tells you the kernel is a normal subgroup which is a subset of H. If it's just {e} then you know it's injective. If it's not {e} then it must be all of H. That would tell you H is normal in G. If it is then use your previous results to draw the conclusion that G has a nontrivial center.

    Now tell me why, if you know G isomorphic to a subgroup of S4, that that subgroup must be A4? Also what they told you is that 2 doesn't divide o(C(G)). That means o(C(G)) could be 1 (trivial), but that means it could also be 3, right? So now I guess you have to show there is no group of order 12 such that o(C(G))=3. That's seems unrelated to the first problems and is annoying. I refuse to think about it. Can you handle that one for me?
     
  18. Dec 10, 2009 #17
    Well... I've reached the same results as you ...
    Now... C(G) is always a normal Sub-Group ... and by the contradiction we get from the previous parts of the question, we can tell that C(G) must be of order 1 (because there isn't a normal subgroup of order 3! ) ...

    I'll try to handle it on my own... Tnx a lot for your help...
    BTW-it isn't related to the first problems as I told you...
     
  19. Dec 10, 2009 #18

    Dick

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    Oh, yeah. If o(C(G))=3 then the first part tells you there another element in the center. Cute. Not such a 'separate problem' as I thought.
     
  20. Dec 10, 2009 #19
    So will you help me continue now? :)
     
  21. Dec 10, 2009 #20

    Dick

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    Sure. What parts do you still need to prove?
     
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