Why $\mathbb{Z}/3\mathbb{Z} \cong \mathbb{Z}_{3}$

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

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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
This is a common definition of $\mathbb{Z}_3$. What definition are you using? – Chris Eagle Mar 21 '12 at 18:49
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

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.

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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 $\mathbb{Z}/3\mathbb{Z}$.

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