Consider a regular octagon with vertices labeled cyclically. Consider $C_8 \hookrightarrow D_8 \hookrightarrow S_8$, where the copy of $C_8$ is generated by a counterclockwise rotation $r$, the copy of $D_8$ is all the (rigid) symmetries of the octagon, and the symmetric group is given by all the permutations of the labels, whether or not they can be rigidly achieved.
a) What is $|D_8/C_8|$? What does an arbitrary element of $D_8/C_8$ look like? Name two distinct elements. Is $D_8/C_8$ a group?
b) What is $|S_8/D_8|$? What does an arbitrary element of $S_8/D_8$ look like? Name two distinct elements. Is $S_8/D_8$ a group?
My attempt:
a) As I understand $C_8=\{e,r,r^2,...,r^7\}, D_8=\{e,r,r^2,...,r^7,s,sr,...,sr^7\}.$ Then $D_8/C_8=\{C_8, D_8\setminus C_8\},$ so $|D_8/C_8|=2. D_8\setminus C_8 =sC_8=\{s,sr,...,sr^7\}.$ Next, $D_8/C_8$ is a group if and only if $C_8$ is normal. But I don't know how to prove/disprove that...
b) I'm completely stuck here. How to write down the cosets of $D_8$ in $S_8$?
$C_n$ is a subgroup of $D_n$ of index $2$ for all $n\ge 3$, and hence $C_n$ is a normal subgroup of $D_n$ with $D_n/C_n\cong C_2$. So it has $2$ elements. The subgroup $D_8$ however is not normal in $S_8$, as mentioned by Arturo, but the quotient set has $\frac{|S_8|}{|D_8|}=8!/16=2520$ elements.