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    So for C6 (cyclic group)

    {1,x,x^2,...,x^5}, you can easily determine the generators by just taking the coprime powers. So x and x^5 in this case.

    But if I had a composition that was isomorphic to C6 ie:

    g1(w) = w
    g2(w) = w^2
    .
    .
    .
    g6(w) = w^6

    What would be the quickest method to find the generators now?

    I worked it out (the long way), and the generators are g3(w) = w^3 and g5(w) = w^5. I would have thought, take powers coprime to 6, again, but clearly its not the case, as 3 is not coprime to 6.

    Thanks

    EDIT: the reason im asking this is because I am currently working on C16 with:

    g1(w) = w
    g2(w) = w^2
    .
    .
    .
    g16(w) = w^16

    And you can see, that it would help alot if I knew a trick to find the generators

    Thanks

    EDIT 2: I think I have found the solution

    so gi would be a generator of C16 if i^(16) mod 17 = 1. But I get quite big numbers for this. Ill wait for some more suggestions

    EDIT3: Actually my solution is not correct. Because i^16 mod 17 can be 1, but not a generator. Sheesh
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    If you have a group that is isomorphic to C_6, just write down the isomorphism explicitly and then the generators will correspond to generators and you can use the usual method for identifying generators of a cyclic group.

    I don't quite understand your notation in your example.

    If w is just some number whose sixth power is 1 (and all smaller powers aren't one) then you will have made a mistake... w^6 would be one and the generator would be w.
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    (Original post by Mark85)
    If you have a group that is isomorphic to C_6, just write down the isomorphism explicitly and then the generators will correspond to generators and you can use the usual method for identifying generators of a cyclic group.

    I don't quite understand your notation in your example.

    If w is just some number whose sixth power is 1 (and all smaller powers aren't one) then you will have made a mistake... w^6 would be one and the generators would be w and w^5.
    Sorry forgot to mention in the C6, w is the 7th root of unity. So w^7 = 1, and respectively in C16, w^17 = 1

    so i have:

    g1(w) = w
    g2(w) = w^2
    .
    .
    .
    g6(w) = w^6

    Where {g1,g2,g3,g4,g5,g6} form C6. But I dont nkow a quick method of working out the generator
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    (Original post by 2710)
    Sorry forgot to mention in the C6, w is the 7th root of unity.
    Here is your mistake, if w is a seventh root of unity, then w,w^2,...,w^6 isn't even a group unless w isn't primitive in which case some of the w^i will be equal thus the total number of them will be less than six.

    Presumably, you want w to be a primitive root of unity in which case there are seven of them (remember w^7=1)

    Also, note that primitive nth roots of unity are precisely those that generate the cyclic group of nth roots.

    So if w is a primitive seventh root of unity, and you take the group 1,w,w^2,...,w^6

    then the obvious isomorphism to 1,x,...,x^6 is

    w-->x and you can read off the generators like that.
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    (Original post by Mark85)
    Here is your mistake, if w is a seventh root of unity, then w,w^2,...,w^6 isn't even a group unless w isn't primitive in which case some of the w^i will be equal thus the total number of them will be less than six.
    g1(w) = w
    g2(w) = w^2
    .
    .
    g6(w) = w^6

    Take g3 = x

    x(w) = w^3
    x2(w) = w^9 = w^2
    x3(w) = w^6
    x4(w) = w^18 = w^4
    x5(w) = w^12 = w^5
    x6(w) = w^15 = w

    Here x2(w) represents x(x(w))

    So w,...,w^6 is a group isomorphic to C6?

    EDIT. I think we are misunderstanding. I am not forming a group w,...,w^6 with the operation multiplication. I am forming the group under composition of functions. I think...

    EDIT: If I give you the full context. The gi's are the elements of the Galois group I am trying to find. So they are the Q automorphisms, which send w to w^2 or w^3 or...w^6
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    (Original post by 2710)
    g1(w) = w
    g2(w) = w^2
    .
    .
    g6(w) = w^6

    Take g3 = x

    x(w) = w^3
    x2(w) = w^9 = w^2
    x3(w) = w^6
    x4(w) = w^18 = w^4
    x5(w) = w^12 = w^5
    x6(w) = w^15 = w

    Here x2(w) represents x(x(w))

    So w,...,w^6 is a group isomorphic to C6?

    EDIT. I think we are misunderstanding. I am not forming a group w,...,w^6 with the operation multiplication. I am forming the group under composition of functions. I think...

    EDIT: If I give you the full context. The gi's are the elements of the Galois group I am trying to find. So they are the Q automorphisms, which send w to w^2 or w^3 or...w^6
    Ok, I see. Your Galois group is the group of units of the integers mod 7. A generator in there is called a primitive root and in general, there is no easy way to find one other than by brute force. This shouldn't be too bad with small examples and wikipedia gives the following method for larger ones: http://en.wikipedia.org/wiki/Primiti...rimitive_roots
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    (Original post by Mark85)
    Ok, I see. Your Galois group is the group of units of the integers mod 7. A generator in there is called a primitive root and in general, there is no easy way to find one other than by brute force. This shouldn't be too bad with small examples and wikipedia gives the following method for larger ones: http://en.wikipedia.org/wiki/Primiti...rimitive_roots
    Ok thank you!
 
 
 
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