Interpretation of homomorphism question

  • #1
Hey,

I had this question I thought i answered but now I'm questioning if the group operation is supposed to be addition and not multiplication.

The question [itex]\varphi :\mathbb{Z}\to {{\mathbb{Z}}_{n}}\,\,\,\,\,where\,\,\varphi (a)=\text{Remainder}\left( \frac{a}{n} \right)[/itex]

When I initially did it, I just mindlessly took the group operation to be multiplication where as it could be addition but it is not directly stated.

So do you guys think it would be multiplication to give
[itex]\varphi (a)\varphi (b)=\text{Remainder}\left( \frac{a}{n} \right)\times \text{Remainder}\left( \frac{b}{n} \right)[/itex]

or addition to give

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

Either way its not a homomorphism but I want understand how it works properly.

Is the general way to write this [itex]\varphi (a)*\varphi (b)[/itex] where * is the binary operation if the group the homomorphism is acting on unless given?

Thanks
 

Answers and Replies

  • #2
jgens
Gold Member
1,581
50
Reduction modulo n is a group homomorphism. Now check the group axioms for Z/nZ under multiplication to see if you have a group. That should answer your question.
 

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