An often given example of a group of infinite order where every element has infinite order is the group $\dfrac{\mathbb{(Q, +)}}{(\mathbb{Z, +})}$.

But I don't see why every element necessarily has finite order in this group. Why is this true?

Also, what is the identity element of this group?

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    $\begingroup$ You have reversed the finite and infinite in what you ask about. Anyway, the identity element is the coset consisting of all the integers. $\endgroup$ – Tobias Kildetoft Feb 11 '16 at 13:02
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    $\begingroup$ It's impossible what you wrote in header. What is possible is an infinite group where every element has finite order. $\endgroup$ – DonAntonio Feb 11 '16 at 13:02

The identity of $G =\mathbb Q / \mathbb Z$ is $\mathbb Z$.

1) Every element of $G$ has finite order. Indeed, if you take any $r + \mathbb Z \in G$ where $r \in \mathbb Q$ then we may write $r = \frac{m}{n}$, with $m,n \in \mathbb Z $ and $n > 0$. Thus

$$n (r + \mathbb Z) = m + \mathbb Z = \mathbb Z$$

then it follows that $\left|r + \mathbb Z\right| \leq n$.


2) The group has infinite order because we may choose an element in $G$ with arbitrarily large order, consider $\frac{1}{n} + \mathbb Z$, where $\left|\frac{1}{n} + \mathbb Z\right| = n$ (why?).


If the group is of finite order then the order of every element in the group divides the order of the group. Hence no element can have infinite order.

In your example, if $\dfrac {p}{q} +\mathbb Z \in \mathbb Q/\mathbb Z$ then $\left(\dfrac{p} {q} +\mathbb Z\right)^q =q\left(\dfrac {p} {q} +\mathbb Z \right) =\mathbb Z$ which is the identity element of this group. Hence every element is of finite order.


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