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Let $n\in \mathbb{Z}^+$. How do I prove that $\mathbb{Z}/n\mathbb{Z}$ is isomorphic to $\mathbb{Z}_n$?

Is there any good homomorphism $\phi$ I could use that graphs $\mathbb{Z}/n\mathbb{Z}$ to $\mathbb{Z}_n$, that is one to one and onto. I'm not sure how to come up with one.

Thank you for any input!

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    $\begingroup$ Divide $m$ by $n$ and take the remainder. $\endgroup$ – user61527 Jan 21 '14 at 19:21
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    $\begingroup$ The most obvious one sends $x$ to $x$. What precise definition are you using for $\mathbb{Z}_n$ anyways? $\endgroup$ – Hurkyl Jan 21 '14 at 19:32
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    $\begingroup$ How do you define $\mathbb{Z}_n$? If some professor told you that $\mathbb{Z}_n := \{0,1,\dotsc,n-1\}$, then good night ... $\endgroup$ – Martin Brandenburg Jan 21 '14 at 19:35
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    $\begingroup$ I thought ${\mathbb Z}_n$ was defined as ${\mathbb Z}/n{\mathbb Z}$. But as a general strategy, if you are trying to prove that $G/N \cong H$, then look for a homomorphism of $G$ onto $H$ with kenrel $N$ and use the First Isomorphism Theorem. $\endgroup$ – Derek Holt Jan 21 '14 at 20:02
  • $\begingroup$ @Derek Holt I was going to say the same thing you did. I should not be surprised that the author of one of the texts I regularly use beat me to it. $\endgroup$ – N. Owad Jan 21 '14 at 20:29
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Hint: Given an integer $m$, use the division algorithm to write $$m = nq + r$$ Define $\phi(m) = r \in \mathbb{Z}_n$ and check the conditions for the first isomorphism theorem (or just show that $\phi$ works directly).

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Hint $\ $ Since in $\,\Bbb Z\,$ one may divide with unique remainder by $\,n,\,$ the coset $\rm\: j +(n) \in \Bbb Z/n\:$ may be uniquely represented by its least element $\ge 0,\,$ the remainder $\rm\:j\ mod\ n\, =\, j - kn.\:$ Thus the set $\,\Bbb Z_n =$ naturals $< n$ form a complete system of representatives of $\rm\,\Bbb Z/n.\,$ Hence we can represent the ring by these "normal forms", and pullback the ring operations to the normal form reps (transport of structure), $ $ e.g. multiplication transported to $\,\Bbb Z_n\,$ becomes $\rm\,\ j * k\, :=\, jk\ mod\ n.\:$

The same remainder representation works for any Euclidean domain with unique remainders, i.e. any domain with a division algorithm with unique smaller remainder. For example, in a polynomial ring $\rm\,K[x]\,$ over a field $\rm\,K\,$ we can divide with unique remainder by any polynomial $\rm\,f,\,$ hence the coset $\rm\: g +(f) \in K[x]/(f) = K[x]\bmod f\:$ may be uniquely represented by its least degree element, the remainder $\rm\:g\ mod\ f\, =\, g - hf.\:$ Therefore the polynomials of $ $ degree $\rm < deg\ f\,$ form a complete system of representatives of $\rm\,K[x]/(f).\,$ Thus we can represent the ring by these normal forms, and pullback ring operations to the normal form reps (transport of structure), e.g. $\rm\: g * h := gh\ mod\ f.\:$

For example, Hamilton's presentation of $\Bbb C$ as pairs of reals is a special case of the above, namely $\:\Bbb R[i]\cong \rm\Bbb R[x]/(\color{#c00}{x^2\!+1}),\:$ with normal forms all linear polynomials $\rm\,(a,b) := {\rm\:a + bx }\:$ with the ring operations transported to the pair normal form reps, e.g. transported multiplication of pairs is

$\rm\begin{eqnarray}\rm (a,\ b) &&\rm (c,\ d) &\!\!=&\rm (ac\!\color{#000}{\bf -}\!bd,\quad\ \ ad\!+\!bc)\\ \rm i.e.\ \bmod{\color{#c00}{\,x^2\!+1}}\!:\ \ \color{#c00}{x^2\equiv -1}\ \,\Rightarrow\,\ (a\! +\! b\color{#c00}x)&&\rm(c\! +\! d\color{#c00}{ x})\, &\!\!\!\rm\,\equiv&\rm (ac\!\color{#c00}{\bf -}\!bd) + (ad\!+\!bc)\, x\\\ \rm i.e.\quad\, (a\! +\! b{\it i}\,)&&\rm(c\! +\! d {\it i}\,)\, &\!\! =\,&\rm (ac\!\color{#000}{\bf -}\!bd) + (ad\!+\!bc)\,{\it i}\end{eqnarray}$

There are multidimensional generalizations of the division algorithm (e.g. Grobner bases) which extend the above to certain multivariate polynomial rings $\rm\,R[x,y,z\ldots]/(f,g,h,\ldots).\:$

The above can be viewed as ring-theoretic special cases of very general methods in term rewriting systems for solving word problems in (quotient) equational algebras, e.g. the Knuth-Bendix completion algorithm. For more on the ring-theoretic perspective see George Bergman's classic paper: The diamond lemma for ring theory, 1978. and its errata and updates. Chasing links to this will locate recent literature on these topics (generalizations of Grobner bases, etc).

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First look at the construction of $\Bbb Z/n\Bbb Z$. Notice that the congruence class is defined as $[0]=\{\cdots,-2n, -n, 0, n, 2n,\cdots\}=n\Bbb Z$. Now because $n\Bbb Z$ is a normal subgroup of $\Bbb Z$, it does not matter what type of coset we construct. Now to construct the set of all cosets of $\Bbb Z$ modulo $n\Bbb Z$ we need to look at what the cosets look like. Now a coset of $m(n\Bbb Z)=\{m+a:a\in n\Bbb Z\}$. Now for $m(n\Bbb Z)$ we can compute it as $\{\cdots,-2n+m,-n+m,0+m,n+m,2n+m,\cdots\}=[m]$ notice how the coset is exactly the definition of a congruence class. Therefore the set of cosets $\Bbb Z/n\Bbb Z$ is directly equal to $\{[0], [1],\cdots,[n-1]\}=\Bbb Z_n$ thus an isomorphism between the two is an automorphism. A trivial automorphism to choose would be the idenity automorphism $\imath:x\mapsto x$.

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