- #1
physicsjock
- 89
- 0
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?
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?