Group homomorphism/Homomorphism theorem/Residue class groups of Z/Example

We consider the surjective group homomorphisms

$\varphi \colon \mathbb {Z} \longrightarrow \mathbb {Z} /(4)$

and

$\psi \colon \mathbb {Z} \longrightarrow \mathbb {Z} /(12).$

We have

${}\operatorname {kern} \psi =\mathbb {Z} \cdot 12\subseteq \mathbb {Z} \cdot 4=\operatorname {kern} \varphi \,.$

Due to fact, there exists a uniquely determined group homomorphism

${\tilde {\varphi }}\colon \mathbb {Z} /(12)\longrightarrow \mathbb {Z} /(4),$

which is compatible with the remainder mappings. The morphism ${}{\tilde {\varphi }}$ sends the remainder of a number after division by ${}12$ to the remainder after division by ${}4$. In particular, the theorem implies that the second remainder does only depend on the first remainder, not on the number itself.

If, to the contrary, we consider

$\varphi \colon \mathbb {Z} \longrightarrow \mathbb {Z} /(5)$

and

$\psi \colon \mathbb {Z} \longrightarrow \mathbb {Z} /(12),$

then

${}\operatorname {kern} \psi =\mathbb {Z} \cdot 12\not \subseteq \mathbb {Z} \cdot 5=\operatorname {kern} \varphi \,,$

and there does not exists a natural mapping

$\mathbb {Z} /(12)\longrightarrow \mathbb {Z} /(5).$

For example, the numbers ${}1,13,25,37,49$ have modulo ${}12$ the remainder ${}1$ but modulo ${}5$ their remainders are ${}1,3,0,2,4$.