# Example of an Infinite Group Whose Elements Have Finite Orders ## Problem 594

Is it possible that each element of an infinite group has a finite order?
If so, give an example. Otherwise, prove the non-existence of such a group. Add to solve later

## Solution.

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. Add to solve later

### 2 Responses

1. Lalitha says:

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

• Yu says:

Close