Consider the short exact sequence
$1 \rightarrow A \xrightarrow{ \phi } G \xrightarrow{ \psi } B\rightarrow 1$
where $\phi(A)=\langle x|- \rangle$ be infinite cyclic group, and $B=\langle \bar{y}| \bar{y}^n\rangle$ be finite cyclic group of order $n$.
($G$ may be abelian or may not be)
Since $G \xrightarrow{ \psi } B$ is surjective, $\exists$ $y\in G$ such that $\psi(y)=\bar{y}$. Then $\psi(y^{n})=\bar{y}^n=1$, so $y^n\in ker(\psi)=\phi(A)$.
Without loss of generality, let $y^{n}=x^{i}$ for some non-negative integer $i$
. It is clear that $G$ is generated by $x,y$.
Also as $A\triangleleft G$, we have $y^{-1}xy\in \phi(A)=\langle x |-\rangle$; so let $y^{-1}xy=x^{j}$ for $j\neq 0$.
Then $y^{-2}xy^{2}=x^{j^2}$, and continuing this $y^{-n}xy^{n}=x^{j^n}$, i.e $x=x^{j^n}$, since $y^n(=x^i)$ commutes with $x$.
So $j^n=1$ in $\mathbb{Z}$. Since $n>0$.
Case1: If $n$ is odd, then $j=1$ is its only solution.
Therefore $y^{-1}xy=x$, which means $G$ is an abelian group; it is quotient of $\mathbb{Z} \oplus \mathbb{Z}$:
$G=\langle x,y | y^{-1}xy=x, y^n=x^i \rangle$
(I didn't find more simplification of this! You may be interested in structure of these groups!)
Case 2: If $n$ is even, then $j=\pm 1$. Case $j=1$ has considered above.
So let $j=-1$. As $y^{-1}xy=x^{-1}$, we have $y^{-1}x^{i}y=x^{-i}$ i.e. $x^{i}=x^{-i}$, since $x^{i}(=y^n)$ commutes with $y$. So $i=0$ i.e. $y^n(=x^i)=1$.
In this case $G$ is semidirect product of $\langle x|- \rangle \cong \mathbb{Z}$, and $\langle y|y^n\rangle \cong \mathbb{Z}/n$.
For this, we have to consider homomorphism from $\mathbb{Z}/n$ into $Aut(\mathbb{Z})\cong \mathbb{Z}_2$. Since $n$ is even, there are only two possibilities of homomorphisms -trivial and $y\mapsto \{x\mapsto x^{-1} \}$.
Therefore $G$ is one of the following:
$G=\langle x,y| y^n, y^{-1}xy=x\rangle \cong \mathbb{Z} \oplus \mathbb{Z}/n$
$G=\langle x,y| y^n, y^{-1}xy=x^{-1}\rangle \cong \mathbb{Z} \rtimes \mathbb{Z}/n$
If you look at the way the direct product is defined, in particular the group operation, it's pretty easy to see that the answer is yes. For we have $(a,b)*(c,d)=(ac,bd)=(ca,db)=(c,d)*(a,b)$.
Thus every direct product of abelian groups is abelian.
Best Answer
First of all, we can reduce to $p$-groups. For any finite abelian group $G$, and prime $p$, let $G_p$ be the set of elements of $p$-power of order. Then $G = \prod_p G_p$. We have $\mathrm{Hom}(G, H) \cong \prod_p \mathrm{Hom}(G_p, H_p)$ and one can show (exercise!) that there is a short exact sequence $0 \to A \to B \to C \to 0$ if and only if there is a short exact sequence $$0 \to A_p \to B_p \to C_p \to 0$$ for every $p$.
Finite abelian $p$ groups are classified by partitions: For the partition $\lambda=(\lambda_1, \lambda_2, \ldots, \lambda_r)$ we write $G(p,\lambda) := \prod_{i} \mathbb{Z}/p^{\lambda_i}$.
Let $\lambda$, $\mu$, $\nu$ be three partitions. The following are equivalent:
There is a short exact sequence $$0 \to G(p, \lambda) \to G(p, \nu) \to G(p, \mu) \to 0$$
The Hall polynomial $g_{\lambda \mu}^{\nu}(p)$ is nonzero.
The Littlewood-Richarsdon coefficient $c_{\lambda \mu}^{\nu}$ is nonzero.
Anyone of the zillion equivalent conditions that you can find in, for example, Knutson and Tao or Zelevinsky
For further discussion of this, see Fulton, especially Section 2.