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Homework Help: Group homomorphism problem

  1. Mar 31, 2012 #1
    Hey I've been working on this question,

    How that the following is a homomorphism
    [tex]\theta :{{D}_{2n}}\to {{D}_{2n}}\,\,\,givenby\,\,\,\theta ({{a}^{j}}{{b}^{k}})={{b}^{k}}\,\,\,[/tex]

    [tex]\theta ({{a}^{j}}{{b}^{k}})\theta ({{a}^{m}}{{b}^{n}})={{b}^{k}}{{b}^{n}}[/tex]


    [tex]\theta ({{a}^{j}}{{b}^{k}}{{a}^{m}}{{b}^{n}})=?[/tex]

    From what it looks like it isn't a homomorphism but I'm not sure how to evaluate the last line,

    Does anyone know how to evaluate it?

    I also have another question which I solved numerically but I'm not sure how to show it algebraically,

    Would anyone know how to show
    [tex]\text{Remaider}\left( \frac{ab}{n} \right)\ne \text{Remainder}\left( \frac{a}{n} \right)\times \text{Remainder}\left( \frac{b}{n} \right) For\,\,a,b,n\in \mathbb{Z}[/tex]

    Or would it be sufficient to just know that its not true?
     
  2. jcsd
  3. Mar 31, 2012 #2
    combine your a's and b's via exponent rules and you will see that you get the same result, assuming [itex] a^jb^ka^mb^n=a^ja^mb^kb^n [/itex]
     
  4. Mar 31, 2012 #3
    How can you assume [itex] a^jb^ka^mb^n=a^ja^mb^kb^n [/itex]?

    The dihedral group doesn't commute so you can't assume that ^ can you?

    I tried using the group property ba=a-1b but it just made it uglier

    I might be able to instead say [itex] b^na^m [/itex] is also an element of the domain so


    [tex]\theta ({{a}^{j}}{{b}^{k}})\theta ({{b}^{n}}{{a}^{m}})={{b}^{k}}{{b}^{n}}[/tex]

    but how can I be sure that the homomorphism would act that way with a element of different arrangement?

    then it would follow
    [tex]\theta ({{a}^{j}}{{b}^{k}}{{b}^{n}}{{a}^{m}})={{b}^{k}}{{b}^{n}}[/tex]
    again assuming the homomorphism simply takes only the b elements, regardless of their arrangement around the a's
     
    Last edited: Mar 31, 2012
  5. Mar 31, 2012 #4

    jgens

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    Notice that [itex]a^kb^ja^mb^n = a^{k-m}b^{j+n}[/itex] and use this to prove the necessary homomorphism property. I promise it works and it does not make things all that messy.

    Your remark on the remainders is not true either. What it [itex]b = kn[/itex]?
     
  6. Mar 31, 2012 #5
    Are you sure that is true? Did you mean [itex](ab)^{-1}=b^{-1}a^{-1}[/itex]?


    the claim says [itex]For\ a,b,n\in\mathbb{Z}[/itex] which I would take to imply that any choices are valid. In that case, the claim is correct. A single counter-example is all that is needed to disprove a claim, but if it is homework, the prof. might want a little more explanation.
     
  7. Mar 31, 2012 #6

    jgens

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    It is true that [itex]ab=ba^{-1}[/itex] in Dihedral groups. The group in question is [itex]D_{2n} = \langle a,b | a^n = b^2 = e, ab = ba^{-1} \rangle[/itex]. If you need to review the group, you can read about it here: http://en.wikipedia.org/wiki/Dihedral_group

    The claim is not correct. As you already noted, a single counter-example is all that is needed to disprove the claim and choosing b = kn does just that.
     
  8. Mar 31, 2012 #7
    [itex]b[/itex] can only be a reflection? If [itex]a[/itex] and [itex]b[/itex] can be any element of the group, the statement is not correct. If the elements have been restricted, this is a convention that was not used in my algebra classes. I just wanted to make sure that the question had been explicit since this is a homework section.

    you are correct on the second one. I was reading it as saying that the positive was not true in general, but that is not what the statement was. sorry.
     
    Last edited: Mar 31, 2012
  9. Mar 31, 2012 #8

    jgens

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    Your mapping will not be well-defined if you do not fix meanings for [itex]a,b[/itex]; in particular, one must be a reflection and one a rotation otherwise you cannot write every element of the group in the desired form. Once [itex]a,b[/itex] have been fixed, it follows that [itex]b[/itex] has to be the reflection if the map is to be a group homomorphism.

    Edit: I know the OP did not include any of this information, but the point is that you can kind of figure out what has to be what.
     
  10. Mar 31, 2012 #9
    Thanks for all your answers guys

    so to get

    [itex]a^kb^ja^mb^n = a^{k-m}b^{j+n}[/itex]


    [itex]b^ja^m = a^{-m}b^{j}[/itex]

    Thanks heaps!
     
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