Consider the following Group Theory question:
Find the number of elements in each of the factorgroups: $$ \frac{\mathbb{Z}/6\mathbb{Z}}{\left \langle \bar{2} \right \rangle}, \frac{\mathbb{Z}/12\mathbb{Z}}{\left \langle \bar{8} \right \rangle}, \frac{\mathbb{Z}/12\mathbb{Z}}{\left \langle \bar{3} \right \rangle}, \frac{\mathbb{Z}/12\mathbb{Z}}{\left \langle \bar{5} \right \rangle} $$ (We denote by $\left \langle g \right \rangle$ the cyclic subgroup generated by a given element $g$ of the group at hand.)
I am very much in doubt about the meaning of the question and would therefore greatly appreciate some elucidation. Let me take the factorgroup $\frac{\mathbb{Z}/6\mathbb{Z}}{\left \langle \bar{2} \right \rangle}$ as example. I would interpret the question in the following way: $\left \langle \bar{2} \right \rangle = \{0 \mod 6, 2 \mod 6, 4 \mod 6\}$. Let $G = \mathbb{Z}/6\mathbb{Z}$ and $H = \left \langle \bar{2} \right \rangle$. Now because for finite groups (with which we are dealing in this case) we have $\#G = [G:H]\#H \Rightarrow [G:H] = \frac{\#G}{\#H}$ we can calculate $[G : H]$, which I suppose is the required quantity (as clearly $\#G = 6$ and $\#H = 3$). We see then that $[G : H] = 2$.
The others can then be answered similarly. Is this correct?