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Homework Help: A few examples of group homomorphism

  1. May 25, 2010 #1
    1. The problem statement, all variables and given/known data

    The exercise is to find examples of various homomorphisms from/to various groups.

    Those I'm having problems with are:

    a. f: (Q,+) --> (Q^+,*) which is onto.
    b. f: U20 --> Z64 which is 1-to-1.
    c. f: Z30 --> S10 which is 1-to-1.

    2. Relevant equations

    3. The attempt at a solution

    for a. , the only homomorphism i could think which send 0 to 1 is f(x) a^x,
    apparently, f sends many elements of (Q,+) to R instead of (Q^+,*).

    for b. , the only insight i have is that U20 has 8 elements, and Z64 has 8*8 elements ... not to deep, i know ... but maybe is?

    for c. , no insights at all.

    I would appreciate any clues given.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. May 25, 2010 #2
    To get b. and c. out of the way first (then we can come back to a. which is more interesting).

    b. isn't possible if by [itex]U_{20}[/itex] you mean the group of residues relatively prime to 20 under multiplication. This is because, with that meaning, [itex]U_{20}\cong Z_4\times Z_2[/itex] which is not cyclic (prove both, or at any rate the latter!), whereas any subgroup of a cyclic group ([itex]Z_{64}[/itex] in this case) is cyclic (also prove!).

    c. Does (12) commute with (345)? What then is the order of (12)(345). Then find an element of order 30 in [itex]S_{10}[/itex].
    Last edited: May 25, 2010
  4. May 25, 2010 #3
    a. is also not possible. Look at an element [itex]x\in (\mathbb{Q},+)[/itex] such that [itex]f:x\mapsto 2\in(\mathbb{Q}^+,\times)[/itex] (which must exist). What can you say about [itex]x/2\text{?}[/itex]
  5. May 26, 2010 #4

    For a. , I concluded that if f(x) = 2 then f(x/2) = sqrt(2) , with contradiction to that f(x/2) is (positive) rational.

    For b. , proved.

    For c. , I marked s = (1,2)(3,4,5)(6,7,8,9,10) , o(s) = 30
    and for any x in Z30, f(x) = s^x such that IM(f) = <s> , and f(x) is well defined because it's not dependant of the number one chooses to represent the residue x.

    Thanks for your help, Martin ;)
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