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.

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 […]

Group of $p$-Power Roots of 1 is Isomorphic to a Proper Quotient of Itself
Let $p$ be a prime number. Let
\[G=\{z\in \C \mid z^{p^n}=1\} \]
be the group of $p$-power roots of $1$ in $\C$.
Show that the map $\Psi:G\to G$ mapping $z$ to $z^p$ is a surjective homomorphism.
Also deduce from this that $G$ is isomorphic to a proper quotient of $G$ […]

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 $0+\Z$.