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Group homomorphism question

  1. May 4, 2012 #1
    1. The problem statement, all variables and given/known data

    http://img515.imageshack.us/img515/5954/asdaii.jpg [Broken]


    2. Relevant equations

    Y(a)Y(b)= Y(ab)

    Z10 = {1,3,7,9}

    Z8 = {1,3,5,7}


    3. The attempt at a solution


    Y(1)=1

    Y(3)=3

    Y(7)=5

    Y(9)=7 Y(9.7)=Y(3)=3

    Y(x)=(x-1)(x-3) + x works

    Y(7) = 24 + 7 = 7 (since its in Z8)

    Y(9) = 57 = 1

    but how can you find ALL of these homomorphisms?
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. May 4, 2012 #2
    This is confusing. Do you mean Z10, the group of integers mod 10?

    Or Z10*, the group of UNITS mod 10?
     
    Last edited by a moderator: May 6, 2017
  4. May 5, 2012 #3
    I got confused when i first read the question too,

    but if it was the integers under addition then there wouldn't be any homomorphisms due to the difference in dimension between Z10 and Z8.

    So I took the question to mean

    Z10 = {1,3,7,9} this set under multiplication modulo 10

    Z8 = {1,3,5,7} this set under multiplication modulo 8

    Unless the whole point of the question is to explain where there are no homomorphisms..
    (I have a feeling its not that because that seems to easy for past assignment question)

    Is it possible to find ALL homomorphisms of the two groups I wrote above? Or do you think the meaning of ALL homomorphisms means there is NO homomorphisms and hence its Z10 and Z8 under addition modulo's 10 and 8 respectively?
     
  5. May 5, 2012 #4

    I like Serena

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    Hi linda! :)

    A homomorphism does not have to be bijective (that's only required for isomorphisms).
    So your groups should probably use addition.

    If you use the homomorphism property earlier in you calculations, you'll find that are not quite so many possibilities.
    When you choose Y(3) in your example, the rest follows with Y(3^2) and Y(3^3).
     
  6. May 5, 2012 #5
    hey I like Serena,

    I realised that its supposed to be addition modulo, so the groups I wrote before,

    Z10 = {1,3,7,9}

    Z8 = {1,3,5,7}

    are actually the generators to the groups Z10 and Z8 under addition modulo 10 and 8 respectively,

    so in this case, do you mean I should do something like,

    Y(1) = 1, Y(3) = 3, Y(7) = 5, Y(9) = 7

    and

    Y(1*3) = Y(4) = Y(1^4) = (Y(1))^4 = 4

    Y(1)Y(3) = 1*3 = 4

    so the homomorphism works,

    but would this be the only possible homomorphism since the generators are being mapped directly to one another?
     
  7. May 5, 2012 #6

    I like Serena

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    Those are not the generators.
    What happens if you start with 1 and keep adding 1?

    The sets you mention are really the groups under multiplication.
    Moreover, 3 is a generator of Z10* since 3^2=9 and 3^3=7 mod 10.

    You seem to be mixing addition and multiplication.
    What you would have (with addition) is:
    Y(1+1)=Y(1)+Y(1)
     
    Last edited: May 5, 2012
  8. May 5, 2012 #7
    I didn't mix up between addition, i just used * as the binary operation instead of +

    Y(1*3) = Y(4) = Y(1^4) = (Y(1))^4 = 4

    Y(1)Y(3) = 1*3 = 4


    Or using +

    Y(1+3) = Y(4) = Y(1^4) = (Y(1))^4 = 4

    Y(1)+Y(3) = 1+3 = 4

    Arn't those sets of numbers I gave the cyclic generators of the groups?


    I guess if I scratch that, I start again from here

    Z10 = {01,2,3,4,5,6,7,8,9}

    Z8 = {0,1,2,3,4,5,6,7}

    these are the groups under addition,

    So a homomorphism between them could be

    Y(0) = 0

    up to Y(7) = 7

    then maybe since in Z8, 8 = 8-8 = 0

    Y(8) = 0

    and in Z8 9 = 1

    Y(9) = 1

    but then

    Y(1+9)=Y(10) = Y(0) = 0

    and Y(1) + Y(9) = 1 + 1 = 2

    doesn't work as a homomorphism

    Is there an arithmetic error I'm making or is it because of my selection of mappings?

    Y(1+9), I did 1+9 as you would in Z10 and Y(1) + Y(9) = 1+1 <- the 1+1 as you would in Z8
     
  9. May 5, 2012 #8

    Office_Shredder

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    No, the problem is that the map doesn't work. Observe two things:

    1) If I tell you what Y(1) is you can easily calculate Y(x) for every value of x. So we should just study what the possible values of Y(1) are.
    2) The order of Y(1) has to divide ten since it's in Z10. But the order of Y(1) divides eight also (for reasons independent of the fact that it is in Z10, think about why this must be true).
     
  10. May 5, 2012 #9

    I like Serena

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    I see.
    Sorry. :redface:

    Uhh... now I see the confusion.
    Indeed, each of them is a cyclic generator of the group.

    More specifically, Z10 and Z8 are both generated by just 1.
    This is also written as: Z10=<1> respectively Z8=<1>.
    So I would say that "1" is "the" generator of both groups.
    But then, for instance "3" could also be "the" generator of both groups.

    Apparently you already know that a homomorphism is completely determined by the image of its generators.
    In this case it is completely determined by the image of 1.


    Right.


    With the choice Y(1)=1 everything you write follows, including the contradiction.
    So indeed, the mapping Y(1)=1 does not define a homomorphism.

    As Office Shredder remarked, the order of an element must divide the order of the group.
    A direct implication is that if Y(x)=y, that:
    #x | 10
    #y | 8
    #y | #x

    This severely limits your choices for Y(1). :wink:
     
  11. May 5, 2012 #10
    Thanks guys I worked it out!

    =D =D
     
  12. May 5, 2012 #11

    I like Serena

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    Good! :approve:

    So... how many did you find?
     
  13. May 5, 2012 #12
    I got two =D

    by mapping 1 to 0 or 1 to 4

    thanks again!
     
  14. May 5, 2012 #13

    I like Serena

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    You're welcome. :)

    (Sorry I couldn't help you with your qnumber for 3d4d.)
     
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