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Math Help - Homomorphisms

  1. #1
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    Homomorphisms

    How many homomorphisms are there from Z30 onto Z10? How many are there to Z10? Give a formula for any homomorphism you find in the form f(x)= "expression" mod "some number." Be sure to verify that your homomorphism is well defined.

    I'm not really sure how to do this at all.
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  2. #2
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    Re: Homomorphisms

    Hi,
    You need to know that both \mathbb{Z}_{30} and \mathbb{Z}_{10} are cyclic groups. Denote by 1 the element cl(1) in \mathbb{Z}_{30}. Then 1 is a generator of Z30; any element is of the form m\cdot1 for some integer m. So for any homomorphism f from \mathbb{Z}_{30} to any group, f is completely determined by f(1) -- f(m\cdot1)=m\cdot f(1). So for a homomorphism f into \mathbb{Z}_{10}, there are 10 choices for f(1). I leave it to you to show for any x\in\mathbb{Z}_{10}, f(m\cdot1)=mx is well defined. However, for f to be onto, f(1) must be a generator of \mathbb{Z}_{10}. So just ask yourself: how many generators of \mathbb{Z}_{10} are there?
    Last edited by johng; December 2nd 2013 at 05:43 PM.
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  3. #3
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    Re: Homomorphisms

    Thank you! Ok as for generators of Z10.....please don't laugh or hate me if I get this wrong but I THINK there are four generators: 1,3,7,9....?
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  4. #4
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    Re: Homomorphisms

    Hi again,
    Exactly right. For a cyclic group G of order n with G = <x> (multiplicative notation), xk is a generator if and only if k is relatively prime to n.
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  5. #5
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    Re: Homomorphisms

    So if those are my four onto homomorphisms, how would I write them in the f(x)...form?
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  6. #6
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    Re: Homomorphisms

    Hi again,

    Write \mathbb{Z}_{30}=\{0,1,2\cdots,29\} and \mathbb{Z}_{10}=\{0,1,2\cdots,9\}.

    For the homomorphism corresponding to 5, f(x)=5x where the x on the left is an element of \mathbb{Z}_30 and the 5x on the right is the element of \mathbb{Z}_10 computed by "5x mod 10". Example x=17, f(17)=5*16 mod 10=0.

    Maybe the following will help (too many integers in the above):

    Let G=<a> be a multiplicative cyclic group of order 30 with generator a. Then H=<a^3> is the (unique) subgroup of G of order 10. The 4 homomorphism of G onto H are:

    f_1(a^m)=(a^3)^m=a^{3m} for all m\in\mathbb{Z}, a\mapsto a^3

    f_3(a^m)=((a^3)^3)^m=a^{9m} for all m\in\mathbb{Z}, a\mapsto (a^3 )^3=a^9

    f_5(a^m)=((a^3)^5)^m=a^{15m} for all m\in\mathbb{Z}, a\mapsto (a^3 )^5=a^{15}

    f_7(a^m)=((a^3)^7)^m=a^{21m} for all m\in\mathbb{Z}, a\mapsto (a^3 )^7=a^{21}

    If k is any integer,

    f_k(a^m)=((a^3)^k)^m=a^{3km} for all m\in\mathbb{Z}, a\mapsto (a^3 )^k=a^{3k}

    is a homomorphism of G into H. This is onto H only for k\equiv1, 3, 5, 7 (\mod10)
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  7. #7
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    Re: Homomorphisms

    thanks so much!! You have been a big help!
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