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Math Help - Properties of Group Homomorphisms

  1. #1
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    Properties of Group Homomorphisms

    Hello!

    I have a few questions related to applying properties of group homomorphisms.

    1) How many homomorphisms are there from Z_{20} onto Z_{8} ?

    Well, since both are cyclic, each has to map to a generator, so is it a trick question, I think there is only one, determined by \phi(1) = 1

    ----

    2) How many homomorphisms are there from Z_{20} to Z_{8} ?

    Well, gcd(8,20) = 4 so there are 4?

    I am pretty sure this works for the groups of integers under addition modulo n...

    I am not actually certain what properties the groups need to have such that you can count the number of homomorphisms by taking the gcd of the order of the groups.

    Any help/confirmation appreciated! Thank you.
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  2. #2
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    Quote Originally Posted by matt.qmar View Post
    Hello!

    I have a few questions related to applying properties of group homomorphisms.

    1) How many homomorphisms are there from Z_{20} onto Z_{8} ?

    Well, since both are cyclic, each has to map to a generator, so is it a trick question, I think there is only one, determined by \phi(1) = 1


    First, the above does NOT define a group homomorphism( for example, check what happens with

    0=\phi(0)=\phi(20)=20\cdot\phi(1)...!) .


    There's no epimorphism as required but you haven't yet proved it


    ----

    2) How many homomorphisms are there from Z_{20} to Z_{8} ?

    Well, gcd(8,20) = 4 so there are 4?


    I get, for example, 1\to 0\,,\,1\to 4\,,\,1\to 2...and I can't find any more.

    What's missing here and which would help you also with (1)?? Hint: think of orders of elements under homomorphisms.

    Tonio



    I am pretty sure this works for the groups of integers under addition modulo n...

    I am not actually certain what properties the groups need to have such that you can count the number of homomorphisms by taking the gcd of the order of the groups.

    Any help/confirmation appreciated! Thank you.
    .
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  3. #3
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    I take it that an epimorphism is a surjective homomorphism?

    is the hint you are giving me, that the order of an image under the homomorphism must divide the order of the group being mapped into?

    ie, if
    \phi : G \rightarrow H, then |\phi(g)| divides |H|
    is that true? I think so!

    So there are no surjective homomorphisms, because there would need to be an element of order 8 in Z_{20} to map to an element of order 8 in Z_{8} to get the whole group? Is there some homomorphism property which dictates that?

    and there is 3 homomorphisms to Z_{8}, because elements of Z_{20} have order 1,2,4,5,10,20 and elements of Z_{8} have orders 1,2,4, and 8, and those have 3 possible orders in common?!

    Coincidentially?, Euler's phi function of 8 is 3....?

    Is any/some/all of this reasoning valid?

    Thank you!!
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  4. #4
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    I take it that an epimorphism is a surjective homomorphism?

    is the hint you are giving me, that the order of an image under the homomorphism must divide the order of the group being mapped into?

    ie, if
    \phi : G \rightarrow H, then |\phi(g)| divides |H|
    is that true? I think so!

    So there are no surjective homomorphisms, because there would need to be an element of order 8 in Z_{20} to map to an element of order 8 in Z_{8} to get the whole group? Is there some homomorphism property which dictates that?

    and there is 3 homomorphisms to Z_{8}, because elements of Z_{20} have order 1,2,4,5,10,20 and elements of Z_{8} have orders 1,2,4, and 8, and those have 3 possible orders in common?!

    Coincidentially?, Euler's phi function of 8 is 3....?

    Is any/some/all of this reasoning valid?

    Thank you!!
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  5. #5
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    Quote Originally Posted by matt.qmar View Post
    I take it that an epimorphism is a surjective homomorphism?

    is the hint you are giving me, that the order of an image under the homomorphism must divide the order of the group being mapped into?

    ie, if
    \phi : G \rightarrow H, then |\phi(g)| divides |H|
    is that true? I think so!

    So there are no surjective homomorphisms, because there would need to be an element of order 8 in Z_{20} to map to an element of order 8 in Z_{8} to get the whole group? Is there some homomorphism property which dictates that?


    Sure! if \Phi:G\to H is a group hom. and ord(g)=k\,,\,g\in G , then \Phi(g)^k=\Phi(g^k)=\Phi(1)=1\Longrightarrow ord(\Phi(g))\mid k

    and there is 3 homomorphisms to Z_{8}, because elements of Z_{20} have order 1,2,4,5,10,20 and elements of Z_{8} have orders 1,2,4, and 8, and those have 3 possible orders in common?!


    Yeppers


    Coincidentially?, Euler's phi function of 8 is 3....?


    Not, it is \phi(8)=4

    Tonio



    Is any/some/all of this reasoning valid?

    Thank you!!
    .
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  6. #6
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    Thank you!!

    I see that \phi(8) = 4

    Also, for any group homomorphism \Phi

    There are four possibilites:
    \Phi(1) = 0 has order 1 in Z_{8}
    \Phi(1) = 4 has order 2 in Z_{8}
    \Phi(1) = 2 has order 4 in Z_{8}
    \Phi(1) = 6 has order 4 in Z_{8}

    So is that just a coincidence?
    Or is that because gcd(8,20) = 4 = number of distinct homomorphisms

    Anyways, is there a way of knowing that none of these are onto?

    Thanks!
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  7. #7
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    Quote Originally Posted by matt.qmar View Post

    Anyways, is there a way of knowing that none of these are onto?

    Thanks!
    I believe this is what has been said to you before. If \theta:\mathbb{Z}_{20}\to\mathbb{Z}_8 is surjective then \mathbb{Z}_{20}/\ker\theta\cong\mathbb{Z}_8 but note that this would imply that \displaystyle\frac{\left| \mathbb{Z}_{20}\right| |}{|\ker\theta|}=8\implies 8\mid 20!
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  8. #8
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    Right, of course.

    Thank you!

    I am wondering, though,

    is the number of group HM's \Phi : Z_n \rightarrow Z_m given by gcd(n,m)?
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  9. #9
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by matt.qmar View Post
    Right, of course.

    Thank you!

    I am wondering, though,

    is the number of group HM's \Phi : Z_n \rightarrow Z_m given by gcd(n,m)?
    Yes.
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