We give an example of a group of infinite order each of whose elements has a finite order.
Consider the group of rational numbers $\Q$ and its subgroup $\Z$.
The quotient group $\Q/\Z$ will serve as an example as we verify below.

Note that each element of $\Q/\Z$ is of the form
\[\frac{m}{n}+\Z,\]
where $m$ and $n$ are integers.

This implies that the representatives of $\Q/\Z$ are rational numbers in the interval $[0, 1)$.
There are infinitely many rational numbers in $[0, 1)$, and hence the order of the group $\Q/\Z$ is infinite.
On the other hand, as each element of $\Q/\Z$ is of the form $\frac{m}{n}+\Z$ for $m, n\in \Z$, we have
\[n\cdot \left(\, \frac{m}{n}+\Z \,\right)=m+\Z=0+\Z\]
because $m\in \Z$.
Thus the order of the element $\frac{m}{n}+\Z$ is at most $n$.
Hence the order of each element of $\Q/\Z$ is finite.

Therefore, $\Q/\Z$ is an infinite group whose elements have finite orders.

Fundamental Theorem of Finitely Generated Abelian Groups and its application
In this post, we study the Fundamental Theorem of Finitely Generated Abelian Groups, and as an application we solve the following problem.
Problem.
Let $G$ be a finite abelian group of order $n$.
If $n$ is the product of distinct prime numbers, then prove that $G$ is isomorphic […]

Group of Order 18 is Solvable
Let $G$ be a finite group of order $18$.
Show that the group $G$ is solvable.
Definition
Recall that a group $G$ is said to be solvable if $G$ has a subnormal series
\[\{e\}=G_0 \triangleleft G_1 \triangleleft G_2 \triangleleft \cdots \triangleleft G_n=G\]
such […]

The Group of Rational Numbers is Not Finitely Generated
(a) Prove that the additive group $\Q=(\Q, +)$ of rational numbers is not finitely generated.
(b) Prove that the multiplicative group $\Q^*=(\Q\setminus\{0\}, \times)$ of nonzero rational numbers is not finitely generated.
Proof.
(a) Prove that the additive […]

Commutator Subgroup and Abelian Quotient Group
Let $G$ be a group and let $D(G)=[G,G]$ be the commutator subgroup of $G$.
Let $N$ be a subgroup of $G$.
Prove that the subgroup $N$ is normal in $G$ and $G/N$ is an abelian group if and only if $N \supset D(G)$.
Definitions.
Recall that for any $a, b \in G$, the […]

Normal Subgroups, Isomorphic Quotients, But Not Isomorphic
Let $G$ be a group. Suppose that $H_1, H_2, N_1, N_2$ are all normal subgroup of $G$, $H_1 \lhd N_2$, and $H_2 \lhd N_2$.
Suppose also that $N_1/H_1$ is isomorphic to $N_2/H_2$. Then prove or disprove that $N_1$ is isomorphic to $N_2$.
Proof.
We give a […]

Torsion Subgroup of an Abelian Group, Quotient is a Torsion-Free Abelian Group
Let $A$ be an abelian group and let $T(A)$ denote the set of elements of $A$ that have finite order.
(a) Prove that $T(A)$ is a subgroup of $A$.
(The subgroup $T(A)$ is called the torsion subgroup of the abelian group $A$ and elements of $T(A)$ are called torsion […]

Please explain what are representatives in Q/Z ? You mean elements of Q/Z?
“This implies that the representatives of Q/Z are rational numbers in the interval [0,1).”

From this sentence please explain the remaining part once again :
“On the other hand, as each element of Q/Z is of the form mn+Z for m,n∈Z, we have…”
Why m=0? and why is the order of any element in Q/Z is atmost n ?

Please explain what are representatives in Q/Z ? You mean elements of Q/Z?

“This implies that the representatives of Q/Z are rational numbers in the interval [0,1).”

From this sentence please explain the remaining part once again :

“On the other hand, as each element of Q/Z is of the form mn+Z for m,n∈Z, we have…”

Why m=0? and why is the order of any element in Q/Z is atmost n ?

Thanks

Note that each element in the group $\Q/\Z$ is of the form $q+\Z$, where $q\in $\Q$. We say that $q$ is a representative for the element $q+\Z$.

For the second part, $m$ is not necessarily $0$. But $m+\Z$ is equal to $0+\Z$.