The homomorphism associated to congruence modulo n

Question

From my understanding of congruence classes, to each group homomorphism $\phi: G \to G'$, there exists an associated congruence relation: $a \equiv b$ if $\phi(a)=\phi(b)$.

Applying this to congruence modulo n, it seems as if there should exist some homomorphism $\phi: \mathbb{Z} \to \mathbb{Z} /n\mathbb{Z}$ (where the group operation is addition) with $$a \equiv b \mod n \quad \text{if} \quad \phi(a)=\phi(b)$$

Does this homomorphism exist? I have tried the following mapping, though it fails to be a homomorphism:

Let $x \in \mathbb{Z}$, and use division with remainder to write $x = qn+r$ where $q,r \in \mathbb{Z}$ and $0 \leq r < n$. Then, define

$$\phi(x)=\phi(qn+r)=r$$

However, I encounter the following problem: If we let $y \in \mathbb{Z}$ with $y=nq'+r'$, and $r+r'\geq n$, then

$$\phi(x+y) = \phi((qn+r)+(q'n+r')) = r+r'-n$$ But $$\phi(x)+\phi(y)= \phi(nq+r) + \phi(nq'+r') = r+r'$$ So $\phi$ is not a homomorphism. Does anyone have any ideas?

Answer 1

The homomorphism you seek for is defined by \begin{align} \varphi:\mathbf Z&\longrightarrow \mathbf Z/n\mathbf Z, \\ x&\longmapsto [x]=x +n\mathbf Z. \end{align} You can check it is a group homomorphism if you consider the correct definition of addition modulo $n$ (I use your notations): $$[x]+[y]=[r+r']=r+r'+n\mathbf Z.$$ You can even check it is actually a ring homomorphism since $[1]=1+n\mathbf Z\;$ is the multiplicative unit, and $\;[x][y]=[xy]=rr'+n\mathbf Z$.

Answer 2

The homomorphism you're interested in is $\phi:\Bbb Z\to\Bbb Z_n$ given by $\phi (x)=x\pmod n$.

Then $\operatorname {ker}\phi=n\Bbb Z$.

In general, since we are considering a homomorphism $\phi: G\to G'$, the congruence $a\cong b$ if $\phi(a)=\phi (b)$, just means $\phi (ab^{-1})=e$. That is, $ab^{-1}\in\operatorname {ker}\phi$.

The kernel of a homomorphism is always a normal subgroup; and there is a canonical submersion from $s:G\to\dfrac G{\operatorname {ker}\phi}$.

Moreover, this quotient has a universal property: for any homomorphism $\phi: G\to G'$, $\phi$ "factors through $s$". That is, there exists a homomorphism $\bar \phi:\dfrac G{\operatorname {ker}\phi}\to G'$ such that $\phi=\bar\phi\circ s$.

In our case, $\phi$ turns out to be the canonical submersion $s:\Bbb Z\to\Bbb Z_n$.