If G is the quotient group $\mathbb{Q}/\mathbb{Z}$ show that G has exponent $0$. Let G = $\mathbb{Q}/\mathbb{Z}$, and show that G has exponent $0$.
I don't see how $0$ would be the exponent, because if I am understanding the definitions correctly this would say that each coset in G (added to itself $0$ times) would give the identity for G which is $\mathbb{Z}$. 
 A: We want to show that the quotient group has elements of arbitrarily high order. This is immediate, since the equivalence class of $\frac{1}{n}$ has order $n$.
A: The exponent of a group is defined to be the least common multiple of the order of all of the elements. Put another way, it's the smallest positive number $k$ such that $g^k=id_G$ for all $g\in G$.
The problem is that this is not always defined. For example, if there is an element of infinite order. There are also examples of times where there are no elements of infinite order and it still doesn't work. The group of all roots of unity has only elements with finite order, but for every prime $p$ it has an element of order $p$. Since since the primes are unbounded, it follows this group doesn't have a well defined exponent.
To fix this problem, we (depending on convention) define the order of a group to be $0$ (or $\infty$) when such an LCM doesn't exist. But the way you should think of this is as "the LCM doesn't exist". This is extremely similar to the characteristic of a field. "Characteristic $0$" doesn't really mean something analogous to "characteristic $11$" . It just means "not characteristic $p$ for all $p$"
