1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Gold Member

    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

    User Avatar
    Gold Member

    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

    User Avatar
    Gold Member

    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!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Group homomorphism problem
  1. Group Homomorphism (Replies: 1)

  2. Homomorphism of groups (Replies: 2)

  3. Group Homomorphism? (Replies: 3)

Loading...