# Group homomorphism problem

1. Mar 31, 2012

### 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?

2. Mar 31, 2012

### ArcanaNoir

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

3. Mar 31, 2012

### 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

Last edited: Mar 31, 2012
4. Mar 31, 2012

### 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 $b = kn$?

5. Mar 31, 2012

### DrewD

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

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.

6. Mar 31, 2012

### jgens

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 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.

7. Mar 31, 2012

### 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.

Last edited: Mar 31, 2012
8. Mar 31, 2012

### jgens

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.

9. Mar 31, 2012

### 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!

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