Can someone please explain why is $\mathbb{Z}/3\mathbb{Z}\cong\mathbb{Z}_{3}$?
Why $\mathbb{Z}/3\mathbb{Z} \cong \mathbb{Z}_{3}$
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0@ChrisEagle I guess that $\mathbb{Z}_3$ was defined as the numbers $0,1,2$ with addition, either by table or modular addition... – 2012-03-21
2 Answers
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.
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 $\mathbb{Z}/3\mathbb{Z}$.