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? $\endgroup$ – user14972 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

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).


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).


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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