infinite group with exactly $n$ elements of finite order. Given $n\in \mathbb{N}$, I want to construct an infinite group with exactly $n$ elements of finite order.
Attempt:
I think we can take $G=\mathbb{Z}_{n}\times \mathbb{Z}$. $(a,0)\in G$ is of order equal to order of $a$.
Question 1: What are some other examples that I can construct? 
Question 2: Are there some non Abelian groups having exactly $n$ elements of finite order? I guess I can take product of $\mathbb{Z}_n$ with some matrix ring.
 A: For question 1, you can take any torsion-free (possibly nonabelian) group $H$ and set $G$ to be $\mathbb{Z}_n\times H$.  For example, we can take $H$ to be the additive group of any division ring of characteristic $0$.
For question 2, you can take $G=\mathbb{Z}_n\times F_k$, where $F_k$ is a free group with $k$ generators (well, this answers question 1 too).  If $k=1$, you get your example $\mathbb{Z}_n\times\mathbb{Z}$.  If $k>1$, you get nonabelian examples.  If $k$ is an uncountable cardinal, then you even get an example where $G$ is uncountable.  However, it would be interesting if one can find an indecomposable group $G$ with this property.

Here is an example of infinite indecomposable group $G$ with exactly $n$ elements of finite order, where $n>1$ is an odd integer.  Let $n=p_1^{k_1}p_2^{k_2}\cdots p_r^{k_r}$ be a prime factorization of $n$.  Define $G$ to be the semidirect product $\left(\prod\limits_{i=1}^r\mathbb{Z}_{p_i^{k_i}}\right)\rtimes\mathbb{Z}$, whose underlying set of $G$ is simply the product of sets $\left(\prod\limits_{i=1}^r\mathbb{Z}_{p_i^{k_i}}\right)\times \mathbb{Z}$, and the multiplication in $G$ is given by
$$(x,y)\cdot (z,w)=(x+(-1)^yz,y+w)$$
for all $x,z\in\prod\limits_{i=1}^r\mathbb{Z}_{p_i^{k_i}}$ and $y,w\in\mathbb{Z}$.  In other words, we have
$$(x,y)\cdot (z,w)=\Big(\left(x_1+(-1)^yz_1,x_2+(-1)^yz_2,\ldots,x_r+(-1)^yz_r\right),y+w\Big)\,,$$
where $x=(x_1,x_2,\ldots,x_r)$ and $z=(z_1,z_2,\ldots,z_r)$ with $x_i,z_i\in\mathbb{Z}_{p_i^{k_i}}$.
To show that $G$ is indecomposable, we prove by contradiction.  Suppose otherwise that $G$ has a nontrivial (internal) direct product decomposition $H\times K$ for some subgroups $H$ and $K$.  Write $\pi:G\to\mathbb{Z}$ for the canonical projection $(x,y)\mapsto y$, where $x\in \prod\limits_{i=1}^r\mathbb{Z}_{p_i^{k_i}}$ and $y\in \mathbb{Z}$.  For simplicity, write $N$ for $\prod\limits_{i=1}^r\mathbb{Z}_{p_i^{k_i}}$.  Then, $\pi(H)$ is a subgroup of $\mathbb{Z}$.  Let $\pi(H)$ be generated by an integer $q\geq 0$.
If $q$ is odd, then it can be easily seen that $H$ contains the subgroup with the underlying set $N\times q\mathbb{Z}$.  If there are any other elements of $H$, then we can show that $\pi(H)$ contains an element $q'$ such that $0<q'<q$, which is a contradiction.  Therefore, $H$ is precisely $N\times q\mathbb{Z}$ (and so $q>1$, otherwise $H=G$, but we assume that $H$ is a proper nontrivial subgroup of $G$).  This also shows that $K$ is a subgroup of $G$ isomorphic to $G/H\cong\mathbb{Z}_q$, but this shows that $\mathbb{Z}\cong q\mathbb{Z}\times \mathbb{Z}_q$, which is absurd.  Therefore, $q$ must be even, which we assume from now on.
Next, we look at $K$ instead.  Since $\pi(H)+\pi(K)=\mathbb{Z}$, we conclude that $\pi(K)=t\mathbb{Z}$ for some integer $t\geq 0$ such that $\gcd(t,q)=1$.  Because $q$ is even, $t$ must be odd.  Therefore, the argument from the paragraph above works for $K$, if we replace $H$ by $K$, and we get a contradiction once again.  So, the assumption that $G$ is decomposable is false.

There may be an example of an infinite indecomposable group $G$ with exactly $n$ element of finite order, where $n$ is an even positive integer.  I do not know yet how to construct such a group $G$.
Edit.  It turns out that the same construction seems to work for every even integer $n>0$ divisible by $4$, but I have not checked completely.  This construction does not work for $n\equiv 2\pmod{4}$ simply because $-1=1$ in $\mathbb{Z}_2$ (and if you tried, you would get that $\mathbb{Z}_2$ is a direct product factor of $G$).  
