Homomorphism of dihedral group

**Jesssa** · March 31st 2012, 01:44 AM

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?

**Sylvia104** · March 31st 2012, 12:06 PM

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 $0\ne2\times2.$

**Jesssa** · March 31st 2012, 05:15 PM

Thanks Sylvia it definitely helped

**Jesssa** · April 2nd 2012, 03:02 AM

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

**Sylvia104** · April 2nd 2012, 03:41 AM

Originally Posted by Jesssa

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 $a+b\equiv r_1+r_2\mod n.$

**Jesssa** · April 2nd 2012, 04:05 AM

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!

**Deveno** · April 2nd 2012, 12:42 PM

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).

**Sylvia104** · April 3rd 2012, 12:37 AM

In fact

Originally Posted by Jesssa

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.

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