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Math Help - Problem on homomorphism

  1. #1
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    Problem on homomorphism

    Q: Let G, H be cyclic groups, generated by elements x, y. Determine the condition on the orders m,n of x and y so that the map sending x^i --> y^i is a group homomorphism.

    Ans: For m<n a function which sends x^i --> y^i can not be well defined; but for m>or=n there is no such problem.

    So, the required condition is m > or = n.

    AM I RIGHT?

    PLEASE HELP
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  2. #2
    Super Member ILikeSerena's Avatar
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    Re: Problem on homomorphism

    Quote Originally Posted by arindamnaskr View Post
    Q: Let G, H be cyclic groups, generated by elements x, y. Determine the condition on the orders m,n of x and y so that the map sending x^i --> y^i is a group homomorphism.

    Ans: For m<n a function which sends x^i --> y^i can not be well defined; but for m>or=n there is no such problem.

    So, the required condition is m > or = n.

    AM I RIGHT?

    PLEASE HELP
    Hi arindamnaskr!

    Your condition is true but incomplete.

    Suppose we call the map f.
    Then for any i, we have

    x^i = x^{m+i}

    Since a map is required to have unique images, that means:

    f(x^i) = f(x^{m+i})

    What can you deduce from that (using the homomorphism property)?
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  3. #3
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    Re: Problem on homomorphism

    f(x^i)= f(x^m+i) = f(x^m).f(x^i)=y^m.y^i=y^m+i.
    But where are we going from here?
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  4. #4
    Super Member ILikeSerena's Avatar
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    Re: Problem on homomorphism

    Quote Originally Posted by arindamnaskr View Post
    f(x^i)= f(x^m+i) = f(x^m).f(x^i)=y^m.y^i=y^m+i.
    But where are we going from here?
    You should have:

    \displaystyle y^i = f(x^i) \overset{\boxed{|x|=m}}{=} f(x^m \cdot x^i) \overset{\boxed{homomorphism}}{=} f(x^m) \cdot f(x^i) = y^m \cdot y^i

    This will only hold true if y^m = e while at the same time we know that |y|=n.
    What does that mean for m?
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  5. #5
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    Re: Problem on homomorphism

    That means m= pn, where p=1,2,3,.....
    Anything else?

    thanks..
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  6. #6
    Super Member ILikeSerena's Avatar
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    Re: Problem on homomorphism

    Quote Originally Posted by arindamnaskr View Post
    That means m= pn, where p=1,2,3,.....
    Anything else?

    thanks..
    Yes.
    So n must be a divider of m.
    That's it!
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  7. #7
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    Re: Problem on homomorphism

    Quote Originally Posted by ILikeSerena View Post
    Yes.
    So n must be a divider of n.
    That's it!
    Thank you again
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