Question

Show that Zto Z/nZ is a natural epimorphism

Answer 1

Answer on Question #82353 – Math – Abstract Algebra

Question

Show that $Z$ to $Z/nZ$ is a natural epimorphism.

Solution

Not every homomorphism from $Z$ to $Z/nZ$ is a natural epimorphism.

Say, if $n$ is composite, $n = a \cdot b$, and $a, b \neq 1, n$, let's consider $f: Z \to Z/nZ$ which maps every $z \in Z$ to the residue class $[a \cdot z] \in Z/nZ$.

It is not epimorphism, since

$$f(b) = [a \cdot b] = [n] = [0],$$

so $Im(f)$ is cyclic and its order equals $b$, whilst order of $Z/nZ$ equals $n$, so they could not be isomorphic. So $Im(f) \neq Z/nZ$, so $f$ is not epimorphism.

But if $n$ is prime and $f: Z \to Z/nZ$ is not trivial, then $f$ is always epimorphism, because $Im(f) \leq Z/nZ$, and since $n$ is prime, so its only subgroups are $\{0\}$ and $Z/nZ$ itself, and since $f$ is not trivial, then $Im(f) = Z/nZ$, so $f$ is epimorphism.

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