Evaluating Group Homomorphisms and the Remainder Theorem

[physicsjock]

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,

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?

 

[ArcanaNoir]

combine your a's and b's via exponent rules and you will see that you get the same result, assuming a^jb^ka^mb^n=a^ja^mb^kb^n

 

[physicsjock]

How can you assume a^jb^ka^mb^n=a^ja^mb^kb^n?

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 b^na^m is also an element of the domain so \theta ({{a}^{j}}{{b}^{k}})\theta ({{b}^{n}}{{a}^{m}})={{b}^{k}}{{b}^{n}}

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

then it would follow
\theta ({{a}^{j}}{{b}^{k}}{{b}^{n}}{{a}^{m}})={{b}^{k}}{{b}^{n}}
again assuming the homomorphism simply takes only the b elements, regardless of their arrangement around the a's

 

[jgens]

Notice that a^kb^ja^mb^n = a^{k-m}b^{j+n} 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 b = kn?

 

[DrewD]

I tried using the group property ba=a^-1b

Are you sure that is true? Did you mean (ab)^{-1}=b^{-1}a^{-1}?

Your remark on the remainders is not true either. What it b=kn

the claim says For\ a,b,n\in\mathbb{Z} 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.

 

[jgens]

Are you sure that is true? Did you mean (ab)^{-1}=b^{-1}a^{-1}?

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

the claim says For\ a,b,n\in\mathbb{Z} which I would take to imply that any choices are valid. In that case, the claim is correct.

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.

 

[DrewD]

b can only be a reflection? If a and b 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.

 

[jgens]

If a and b can be any element of the group, the statement is not correct.

Your mapping will not be well-defined if you do not fix meanings for a,b; 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 a,b have been fixed, it follows that b 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.

 

[physicsjock]

Thanks for all your answers guys

so to get

a^kb^ja^mb^n = a^{k-m}b^{j+n}


b^ja^m = a^{-m}b^{j}

Thanks heaps!