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Math Help - Algebra Final

  1. #1
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    Algebra Final

    I need some help with these review problems I have for an upcoming final.


    How many homomorphisms are there of Z onto Z?
    How many homomorphisms are there of Z into Z?
    How many homomorphisms are there of Z into Z_2?

    How would you go about approaching this type of problem?

    Are the following groups isomorphic?
    Z_2 X Z_12 and Z_4 X Z_6

    I think these are because the lcm(2,12)=lcm(4,6). Would this be a right assumption?

    Z_8 X Z_10 X Z_24 and Z_4 X Z_10 X Z_40
    I think these are because the lcm(8,10,24)=lcm(4,12,40)
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by matt90 View Post
    I need some help with these review problems I have for an upcoming final.


    How many homomorphisms are there of Z onto Z?
    There are only 2.

    define \theta_1(x) = x for x \in \mathbb{Z}<br />
    and \theta_2 (x) = -x for x \in \mathbb{Z}

    check that these are homomorphisms


    also, they will be the only ones. since any homomorphism has to be of the form \theta (x) = nx for some integer n. this only works with n = +/- 1, otherwise the map would not be onto.

    How many homomorphisms are there of Z into Z?
    from directly above, we can choose n to be any integer here, so there are an infinite amount.

    How many homomorphisms are there of Z into Z_2?
    there are 2.

    \theta (x) = [0]_2 for all x \in \mathbb{Z}

    or define a mapping based on \theta (1) = [1]_2, in which case \theta (x) = \left \{ \begin{array}{lr} \text{[0]} & \mbox{ if } x \mbox{ is even} \\ & \\ \text{[1]} & \mbox{ if } x \mbox{ is odd} \end{array} \right.

    Are the following groups isomorphic?
    Z_2 X Z_12 and Z_4 X Z_6

    I think these are because the lcm(2,12)=lcm(4,6). Would this be a right assumption?
    Z_2 x Z_12 = Z_2 x Z_3 x Z_4

    and Z_4 x Z_6 = Z_4 x Z_2 x Z_3

    these are the same thing, just rearranged. thus they are isomorphic


    Z_8 X Z_10 X Z_24 and Z_4 X Z_10 X Z_40
    I think these are because the lcm(8,10,24)=lcm(4,12,40)
    the orders are different. thus they are not isomorphic
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    There are only 2.

    define \theta_1(x) = x for x \in \mathbb{Z}<br />
    and \theta_2 (x) = -x for x \in \mathbb{Z}

    check that these are homomorphisms

    also, they will be the only ones. since any homomorphism has to be of the form \theta (x) = nx for some integer n. this only works with n = +/- 1, otherwise the map would not be onto.

    from directly above, we can choose n to be any integer here, so there are an infinite amount.

    there are 2.
    What is the justification for this form \theta (x) = nx being the only one?
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  4. #4
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    yeah i don't understand that part
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  5. #5
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    oh i made a mistake on the last part it should be Z_12

    Z_8 X Z_10 X Z_24 and Z_4 X Z_12 X Z_40
    I think these are because the lcm(8,10,24)=lcm(4,12,40)
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Isomorphism View Post
    What is the justification for this form \theta (x) = nx being the only one?
    1 is a generator of Z, if i can figure out where it goes, i can figure out where anything goes.

    so i start with some homomorphism f, for which f(1) = n ......... n is just some integer.

    then, f(x) = f( \underbrace{1 + 1 + 1 ... + 1}_{\text{x times}}) =  f(1) + f(1) + \cdots f(1) = nx

    is something wrong with that reasoning?
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  7. #7
    Lord of certain Rings
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    Quote Originally Posted by Jhevon View Post
    f(x) = f( \underbrace{1 + 1 + 1 ... + 1}_{\text{x times}}) =  f(1) + f(1) + \cdots f(1) = nx

    is something wrong with that reasoning?
    Nothing wrong, of course

    I could not figure out the reason, so I asked.

    Thanks
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  8. #8
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    Quote Originally Posted by Isomorphism View Post
    Nothing wrong, of course

    I could not figure out the reason, so I asked.

    Thanks
    In general if a group is generated by \left< a_1, a_2, ... , a_k\right>. Then any homomorphism can be determined by \phi(a_1), ... ,\phi(a_k).
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