Every inﬁnite abelian group has at least one element of inﬁnite order? Is the statement true?

Every inﬁnite abelian group has at least one element of inﬁnite order.

I am searching for an infinite abelian group with all elements having finite order.
Please help me to find such groups.
 A: Take an infinite number of generators, and specify that any element squared is the identity.
A: Consider $(\mathbb{Q},+)$ and the subgroup $(\mathbb{Z},+)$. Now take the quotient $\mathbb{Q}/\mathbb{Z}$.
A: Define the set $G$ to be the collection of all functions $f:\mathbb{N}\to \mathbb{Z}/2\mathbb{Z}$. We can then define $f+g$, for any two $f,g\in G$, in the natural way. Define $e$ to be the map such that $e(n) = 0$ for all $n\in \mathbb{N}$. It is not hard to see that $f+f = e$ for all $f\in G$. 
A: There are many counter examples, some of the more natural ones are:


*

*$( \mathbb{N}, \oplus )$ where $\oplus$ is bitwise Exclusive OR of the binary representations between two natural numbers. Every non-zero numbers has order $2$.

*Consider any polynomial rings over any finite field.
If we strip away its multiplication, we get an infinite abelian group with respect to the addition. Every non-zero element has order $p$, the characteristic of the underlying field.
A: Let $G=\{z\in \mathbb{C}:z^n=1$ for some positive integer $n\}$. Then $G$ with complex multiplication is an infinite abelian group but every element has finite order.
A: As mentioned in various answers it is not true that every infinite abelian group contains an element of infinite order; let me try to structure the examples a bit along a standard distinction. 
An abelian group in which every element has finite oder is called a torsion abelian group; more generally, the subsets of elements of finite order form a subgroup called the torsion subgroup. 
Thus what you are looking for are infinite torsion abelian groups. 
A further thing to observe is that even  when the order of each element is finite, it can still happen that there is no common bound for the order of the elements. 
This is the case in, e.g., $\mathbb{Q}/\mathbb{Z}$ and $\oplus_{n\in \mathbb{N}^{\ast}} \mathbb{Z}/n\mathbb{Z}$ (note the direct sum, the direct product would not be a torsion group).  
A torsion abelian group where there is a common bound for the order of the elements is called a bounded abelian groups and the maximal order of an element, which is in fact the least common multiple of all orders, is called the exponent of the group.  
There are also examples of infinite bounded abelian groups, such as $(\mathbb{Z}/n\mathbb{Z})^{\mathbb{N}}$ for some positive integer $n$.  
A: The statement is false
Consider the Power Set of Natural number with group operation of symmetric difference.Then the Group clearly is infinite,abelian,and no element is of infinite order more so each element has order $2$.
A: The most natural example to me is $\prod_{n=1}^\infty\mathbb Z_2$, where every element has order $ 2$.
A: Consider $(\mathbb{F}_p[X],+)$, the polynomial ring over $\mathbb{F}_p$ under addition. This is an abelian infinite group with exponent $p$, which means that $a^p=0$ for every $a \in \mathbb{F}_p[X]$. Since we're looking at "$+$"
$a^p$ is actually $p \cdot a$ in this case.
Notice that this is a stronger property, than "every element has finite order".
