Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Can someone please explain why is $\mathbb{Z}/3\mathbb{Z}\cong\mathbb{Z}_{3}$?

share|improve this question
4  
What do you mean by $\mathbb{Z}_3$? The $3$-adic integers, or the cyclic group of integers under addition modulo $3$? –  Arturo Magidin Mar 21 '12 at 18:49
2  
This is a common definition of $\mathbb{Z}_3$. What definition are you using? –  Chris Eagle Mar 21 '12 at 18:49
2  
Did you learn the first isomorphism theorem? –  N. S. Mar 21 '12 at 18:51
    
@ChrisEagle I guess that $\mathbb{Z}_3$ was defined as the numbers $0,1,2$ with addition, either by table or modular addition... –  N. S. Mar 21 '12 at 18:52

2 Answers 2

up vote 7 down vote accepted

If $\mathbb{Z}_3 = \{\overline{0},\overline{1},\overline{2}\}$ with addition modulo $3$, consider the homorphism $f\colon\mathbb{Z}\to\mathbb{Z}_3$ given by $f(a) = \overline{a\bmod 3}$; that is, $a$ is mapped to its remainder modulo $3$.

Since $(a+b)\bmod 3 = \Bigl((a\bmod 3) + (b\bmod 3)\Bigr)\bmod 3$, $f$ is a group homomorphism. The kernel of $f$ is precisely the elements that are multiples of $3$, i.e., $3\mathbb{Z}$. And $f$ is onto.

By the First Isomorphism Theorem, we have $$\mathbb{Z}_3 = \mathrm{Im}(f) \cong \frac{\mathbb{Z}}{\mathrm{Ker}(f)} = \frac{\mathbb{Z}}{3\mathbb{Z}}.$$

The explicit isomorphism takes $a+3\mathbb{Z}$ to $\overline{a\bmod 3}$; verify that it is well defined and a group homomorphism.

share|improve this answer

I'm going to assume that by $\mathbb{Z}_3$ you mean the set $\{0,1,2\}$ with addition defined mod 3. And I'm going to assume that for $\mathbb{Z}/3\mathbb{Z}$ you have in mind cosets: $\{\{\ldots,-3,0,3,\ldots\}, \{\ldots,-2,1,4,\ldots\}, \{\ldots,-1,2,5,\ldots\}\}$. In each case there are three elements $A$, $B$, and $C$, and the addition table is the same table.

You should know that $\mathbb{Z}_3$ sometimes refer to the $3$-adic integers, which are altogether different. And that even in your context, many mathematicians define $\mathbb{Z}_3$ to be $\mathbb{Z}/3\mathbb{Z}$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.