This exercise asks to find all abelian groups that fit in the short exact sequence
$0\to \mathbb{Z}\to A\to \mathbb{Z}_n\to 0$
I've proved that $A$ must be isomorphic to $\mathbb{Z}\oplus \mathbb{Z}_d$ with $d|n$, but I've been unable check that this group fits.
I have to define an injective homomorphism $\phi:\mathbb{Z}\to \mathbb{Z}\oplus \mathbb{Z}_d$. I guess that $\phi(1)=(1,x)$, where $x$ must be something that makes $\mathrm{coker}(\phi)\cong\mathbb{Z}_n$. In the solutions I've read $x=n/d$, so $\phi(1)=1\cdot (1,0)+n/d\cdot(0,1)$ but I can't show that that makes what I want. A presentation of $\mathrm{coker}(\phi)=\langle a,b\mid db=0, a+bn/d=0\rangle$. If I multiply by $d$, $da=0$, so this is actually a subgroup of $\mathbb{Z}_d\oplus\mathbb{Z}_d$, which has no element of order $n$ in general, and therefore cannot be isomorphic to $\mathbb{Z}_n$.
What am I doing wrong o what should I take as $x$?
Best Answer
Let's define the surjection $\pi:\Bbb Z\oplus\Bbb Z_d\to\Bbb Z_n$ instead. Write $n=cd$, and define $\pi(x,y)=x+cy$. The kernel is generated by $(c,-1)$ and is isomorphic to $\Bbb Z$.