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Thread: Abstract Math Group

  1. #1
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    Abstract Math Group

    1) Find all the distinct generators of the group $\displaystyle \mathbb{Z}_{11}$ w.r.t. multiplication.

    2) Find all the subgroups of group $\displaystyle \mathbb{Z}_{11}$


    Is there a way to find all the distinct generators without computing every element in $\displaystyle \mathbb{Z}_{11}$?
    Last edited by TitaniumX; Mar 18th 2009 at 02:30 PM. Reason: fix latex
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  2. #2
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    Hi

    Of course there is a way: let $\displaystyle n$ be an integer, the generators of $\displaystyle \mathbb{Z}_n$ are the classes of integers which are relatively prime with $\displaystyle n,$ i.e. $\displaystyle \{[x]\in\mathbb{Z}_n;\ gcd(x,n)=1\}$

    That comes from Bezout relation: $\displaystyle gcd(x,n)=1\Rightarrow \exists u,v\in\mathbb{Z},\ ux+vn=1 \Rightarrow [u][x]=[1]$ so $\displaystyle [x]$ is a generator.

    Conversely, $\displaystyle [x]$ generator $\displaystyle \Rightarrow \exists u\in\mathbb{N},\ [u][x]=1\Rightarrow \exists u,v\in \mathbb{Z},\ ux=1-vn$
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  3. #3
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    Quote Originally Posted by clic-clac View Post
    Hi

    Of course there is a way: let $\displaystyle n$ be an integer, the generators of $\displaystyle \mathbb{Z}_n$ are the classes of integers which are relatively prime with $\displaystyle n,$ i.e. $\displaystyle \{[x]\in\mathbb{Z}_n;\ gcd(x,n)=1\}$

    That comes from Bezout relation: $\displaystyle gcd(x,n)=1\Rightarrow \exists u,v\in\mathbb{Z},\ ux+vn=1 \Rightarrow [u][x]=[1]$ so $\displaystyle [x]$ is a generator.

    Conversely, $\displaystyle [x]$ generator $\displaystyle \Rightarrow \exists u\in\mathbb{N},\ [u][x]=1\Rightarrow \exists u,v\in \mathbb{Z},\ ux=1-vn$
    Does the GCD=1 work for the group $\displaystyle {Z}_{11}$ with respect to multiplication?

    Because since 11 is prime, every number less than 11 must also be relatively prime, but not all those numbers are distinct generators. Those are [2],[6],[7],[8]. Can you use some trick to figure those out without testing all the elements in $\displaystyle {Z}_{11}$?
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  4. #4
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    Multiplication? I don't understand, $\displaystyle \mathbb{Z}_{11}$ is an additive group, and since $\displaystyle 11$ is a prime, every non-zero class is a generator.
    Last edited by clic-clac; Mar 18th 2009 at 12:33 PM. Reason: cor
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  5. #5
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    Quote Originally Posted by clic-clac View Post
    Multiplication? I don't understand, $\displaystyle \mathbb{Z}_{11}$ is an additive group, and since $\displaystyle 11$ is a prime, every non-zero class is a generator.
    $\displaystyle {Z}_{11}$ can be a multiplicative group also. All the non-zero elements of $\displaystyle {Z}_{11}$ ([1],[2],[3],[4].....[11]) also form a group under multiplication. This can be verified using a Cayley table. In fact, for any $\displaystyle {Z}_{n}$ when n is prime, it forms a group w.r.t. multiplication.

    It can easily be verified that [2],[6],[7],[8] are distinct generators for the group under multiplication. But I was wondering if there's an easier way to get those answers, like the ones for the additive group just by knowing which element is relatively prime to n?


    EDIT:
    For multiplication, for example [10] only generate the set {[1],[10} and [3] generates the set {[1],[3],[9],[5],[4]} and etc. But the classes I mentioned above will generate the whole group. I got those values by doing it manually, but if n is large like $\displaystyle {Z}_{97}$, it will be a very laborious process.
    Last edited by TitaniumX; Mar 18th 2009 at 02:26 PM.
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  6. #6
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    Quote Originally Posted by TitaniumX View Post
    ([1],[2],[3],[4].....[11])
    Is the last [10]?

    If yes ok, I see that's what I call $\displaystyle \mathbb{Z}_{11}^*$
    This group is isomorphic to $\displaystyle (\mathbb{Z}_{\phi(11)},+)=(\mathbb{Z}_{10},+)$

    So it's a cyclic group, and since only 1,3,7,9 are relative primes with 10, it has 4 distinct generators.

    To easily find the generators, maybe there are formulas but I don't see one now.
    What can be done is, if you know a generator of $\displaystyle \mathbb{Z}_{11}^*$, for instance $\displaystyle [2],$ then the group isomomorphism $\displaystyle \psi\mathbb{Z}_{10},+)\rightarrow (\mathbb{Z}_{11}^*,.)$ such that $\displaystyle [1]\mapsto [2]$ gives you the others generators: $\displaystyle \psi([3]),\ \psi([7]),\ \psi([9])$
    Indeed, $\displaystyle \psi([3])=[2]^3=[8],\ \psi([7])=[8]^2[2]=[-3]^2[2]=[7]$ and $\displaystyle \psi([9])=[8]^3=[-3]^3=[-27]=[6]$
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