An epimorphism from $\mathbb Z_n$ to $\mathbb Z_m$ exists iff $m\mid n$.

Question

I want to show intuitively that for an epimorphism to exist from $\mathbb Z_n$ to $\mathbb Z_m$, $m$ must be a divisor of $n$. Since $f$ is a surjection, so every element in $\mathbb Z_m$ generates a fibre in $\mathbb Z_n$, i.e. a collection of elements in domain that map to a specific element of codomain, note that the fibre $f^{-1}(0_m)$ is the kernel of the homomorphism and the other fibres are simply the cosets of $\ker f$ in $\mathbb Z_n$, now since these cosets are equal in size and disjoint, so each coset contains $\frac{|\mathbb Z_n|}{|\mathbb Z_m|}$ elements, but it must be an integer, so $m$ must divide $n$. Also we can construct easily an epimorphism if $m\mid n$. Is my intuition correct?

Answer 1

This can be seen in more than one way. For instance:

• The image of a cyclic group $G$ under a morphism $f:G\rightarrow H$ must be itself cyclic. In fact if $g$ is a generator of $G$ it is straightforward to check that $f(G)$ is generated by $f(g)$. Now, if $G$ is finite of order $n$, then $n\cdot g=1_G$ and so $n\cdot f(g)=f(n\cdot g)=f(1_G)=1_H$. Hence the order of $f(G)$ divides $n$.

• If $f:{\Bbb Z}_n\rightarrow{\Bbb Z}_m$ is a surjective morphism $f(\bar 1)=\bar a$ must be a generator of ${\Bbb Z}_m$. But $\bar1=\overline{n+1}$ so we need to have $m$ be a divisor of $(n+1)a-a=na$ and so $m$ divides $n$ since ${\rm gcd}(a,m)=1$.

• Given a morphism $f:G\rightarrow H$ we know that $f(G)\simeq G/\ker(f)$. If $G$ is finite the argument leadng to Lagrange's theorem shows that $|G|=|f(G)|\cdot|\ker(f)|$.