The elements of $\mathbb{Z}_{60}/\langle \overline{6}\rangle$ are $\langle \overline{6}\rangle$, $\overline{1}+\langle \overline{6}\rangle$, $\overline{2}+\langle \overline{6}\rangle$, $\overline{3}+\langle \overline{6}\rangle$, $\overline{4}+\langle \overline{6}\rangle$, and $\overline{5}+\langle \overline{6}\rangle$. which if these is equal to $\overline{8}+\langle \overline{6}\rangle$? How many times do you have to add that one to itself to get $\langle \overline{6}\rangle$?
There’s no law against doing some actual computations to see what’s going on!
Remember, $a^n$ in this context doesn't (necessarily) refer to $a$ to the power of $n$, in the traditional sense. What it means is $a * a * a * \ldots * a$, $n$ times, where $*$ is whatever the group operations is. In the case where $*$ is multiplication, then this becomes $a^n$ in the traditional sense. When $*$ is addition, this refers to $a + a + a + \ldots + a = na$.
So, if we take an arbitrary rational number $q \in \mathbb{Q}$, then what's its order in the group under addition? We are solving for $nq = 0$ (as $0$ is the additive identity) for smallest $n \ge 1$.
Most of the time, this has no solution. In particular, when $q \neq 0$, then $n = 0$ is the only solution. Since there is no positive integer $n$ such that $nq = 0$, this means $q$ has infinite order.
On the other hand, when $q = 0$, then any $n$ will do, so the smallest $n \ge 1$ is $n = 1$. Thus, the order is $1$ (as it always is for the identity).
What about for $\mathbb{Q}^*$? I'll leave it to you. I'll give you a hint: there are exactly two elements of finite order.
Best Answer
Let $a = 2/3 + \mathbb{Z}$. Since $3a = 0$ and $a \neq 0$, the order of $a$ is $3$.