Finite cyclic $\Bbb Z$-module and exact sequence

Question

Suppose $M$ is $\Bbb Z$-module, cyclic, finite. How to prove $\require{AMScd}$ \begin{CD} 0 @>>> \Bbb Z @>>> \Bbb Z @>>> M @>>>0\\ \end{CD} is short exact sequence?

in fact we must prove : $\require{AMScd}$ \begin{CD} 0 @>f>> \Bbb Z @>g>> \Bbb Z @>h>> M @>k>>0\\ \end{CD} $f,g,h,k$ exist s.t. $Im f=Kerg, Img=Kerh,Imh=Kerk$

in other way: $g,h$ exist s.t $g$ is one-one function,$h$ is onto function, $Img=Kerh$

if $g=i$ ,$g$ is one-one function, $h=?$,$M=a \Bbb Z$, $h$ is onto but $Img\not =Kerh$

Answer 1

You probably want to prove that

if $M$ is a finite cyclic $\mathbb{Z}$-module, then there exists a short exact sequence $$ 0\to \mathbb{Z}\to\mathbb{Z}\to M\to 0 $$

Let $x$ be a generator of $M$ and consider the map $f\colon\mathbb{Z}\to M$ defined by $f(n)=nx$. This is a surjective homomorphism and its kernel is of the form $\ker f=k\mathbb{Z}$, with $k>0$.

Surjectivity follows by definition of cyclic module and the fact that $x$ is a generator. The homomorphism theorem says that $\mathbb{Z}/k\mathbb{Z}\cong M$, so $k=|M|$. We, of course, use the fact that a subgroup of $\mathbb{Z}$ is of the form $r\mathbb{Z}$, for a unique $r\in\mathbb{N}$, and that, for $r>0$, $\mathbb{Z}/r\mathbb{Z}$ has $r$ elements.

Consider now the map $\mu_k\colon\mathbb{Z}\to\mathbb{Z}$ defined by $\mu_k(n)=nk$.

It is obviously a homomorphism; moreover it is injective, because $k>0$ and so, from $\mu_k(n)=0$ we deduce $n=0$.

Is the sequence $$ 0\to \mathbb{Z}\xrightarrow{\mu_k}\mathbb{Z}\xrightarrow{f} M\to 0 $$ exact?

Yes; $\mu_k$ is injective and $f$ is surjective. Moreover, the image of $\mu_k$ is clearly $k\mathbb{Z}$ which is equal to $\ker f$.