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Math Help - Homomorphism of dihedral group

  1. #1
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    Homomorphism of dihedral group

    Hey I've been working on this question,

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

    \theta ({{a}^{j}}{{b}^{k}})\theta ({{a}^{m}}{{b}^{n}})={{b}^{k}}{{b}^{n}}


    \theta ({{a}^{j}}{{b}^{k}}{{a}^{m}}{{b}^{n}})=?

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

    It would be easy if the group commuted..

    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
    \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}

    Or would it be sufficient to just know that its not true?
    Last edited by Jesssa; March 31st 2012 at 06:47 AM.
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  2. #2
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    Re: Homomorphism of dihedral group

    The dihedral group D_{2n} consists of n rotations and n reflections; the rotations form a cyclic subgroup <a> of order n. An element of D_{2n} can be written as a^jb^k where j\in\mathbb Z, b is any reflection and k=0 or 1.

    A rotation generally does not commute with a reflection; however, any product of a reflection followed by a rotation is equal to a product of a rotation followed by another reflection. In other words, ba^m=a^{m'}b for some other m'\in\mathbb Z. Hope this helps.


    For your second question, if you take a=b=6 and n=4, then 0\ne2\times2.
    Last edited by Sylvia104; March 31st 2012 at 12:09 PM.
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    Re: Homomorphism of dihedral group

    Thanks Sylvia it definitely helped
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    Re: Homomorphism of dihedral group

    Sorry to bug you again,

    I thought I had the question down but then I realised I made an assumption that the group binary operation was multiplication,

    The question was show the following is a homomorphism
    \varphi :\mathbb{Z}\to {{\mathbb{Z}}_{n}}\,\,\,\,\,where\,\,\varphi (a)=\text{Remainder}\left( \frac{a}{n} \right)

    When I did this step I assumed multiplication,

    \varphi (a)\varphi (b)=\text{Remainder}\left( \frac{a}{n} \right)\times \text{Remainder}\left( \frac{b}{n} \right)

    but I now I'm paranoid that it should be addition,

    \varphi (a)+\varphi (b)=\text{Remainder}\left( \frac{a}{n} \right)+\text{Remainder}\left( \frac{b}{n} \right)

    Since in class we mainly use the integers under an addition modulo rather then multiplication, we only do multiplication on integers co-prime to the dimension

    If n = 4 in this case then would the operation be addition modulo ten then 0 = 2+2=4 with a=b=6 like you said before?

    I read online somewhere that "Reduction modulo n is a homomorphism" is that true?

    I found some cases by mucking around in mathematica where its not satisfied under addition, if n=5 a=11 b=8,

    I should of posted the question up in the first place haha

    Sorry again for pestering you once more
    Last edited by Jesssa; April 2nd 2012 at 03:04 AM.
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    Re: Homomorphism of dihedral group

    Quote Originally Posted by Jesssa View Post
    The question was show the following is a homomorphism
    \varphi :\mathbb{Z}\to {{\mathbb{Z}}_{n}}\,\,\,\,\,where\,\,\varphi (a)=\text{Remainder}\left( \frac{a}{n} \right)
    This follows from the fact that if a\equiv r_1\mod n and b\equiv r_2\mod n then a+b\equiv r_1+r_2\mod n.
    Thanks from Jesssa
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    Re: Homomorphism of dihedral group

    Ahh so it is addition!

    phi(a+b) = Rem[(a+b)/n] = Rem[a/n]+Rem[b/n] = phi(a) + phi(b) !

    Thanks again Sylvia!
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    Re: Homomorphism of dihedral group

    here is a (somewhat more) formal proof. suppose a = a' (mod n), and b = b' (mod n). then a+b = a'+b' (mod n).

    since a = a' (mod n), by definition a - a' = kn, for some integer k. we can re-write this as a = a' + kn.

    similarly, since b = b' (mod n), then b - b' = mn, for some integer m. we can re-write this as b = b' + mn.

    therefore: a + b = (a' + kn) + (b' + mn) = a' + b' + (k + m)n. thus (a + b) - (a' + b') = (k + m)n, so

    a + b = a' + b' (mod n) (since k+m is an integer whenever k and m are).

    in particular, if [a] is the unique element of Z congruent to a mod n between 0 and n-1, then [a] + [b] = [a + b],

    which makes the map a-->[a] a homomorphism from Z to Zn: φ(a + b) = [a + b] = [a] + [b] = φ(a) + φ(b).
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    Re: Homomorphism of dihedral group

    In fact

    Quote Originally Posted by Jesssa View Post
    When I did this step I assumed multiplication,

    \varphi (a)\varphi (b)=\text{Remainder}\left( \frac{a}{n} \right)\times \text{Remainder}\left( \frac{b}{n} \right)
    it is also true that \varphi(ab)=\varphi(a)\varphi(b) since it is also true that a\equiv a'\mod n and b\equiv b'\mod n \implies ab\equiv a'b'\mod n. (NB: While 0\ne2\times2, that's only in the ring of integers; it will be true if you reduce modulo 4.) Thus \varphi is a ring homomorphism, not just a homomorphism between additive groups.
    Last edited by Sylvia104; April 3rd 2012 at 12:39 AM.
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