Figuring out what $G/H$ looks like if $G=D_3$ and $H=\{R_0,R_{120},R_{240}\}$

Question

I am having trouble writing out coset elements when they are not representing numbers, such as the dihedral groups.

Question 1: If $G=D_3$ and $H=\{R_0,R_{120},R_{240}\}$, how do you construct $G/H$?

Question 2: And how do you know, after constructing it, that you have exhausted your options? (I am using the fact that every subgroup of $D_n$ consisting of only rotations is normal in $D_n$.

Answer 1

$|D_3| = 6$ so $[D_3 : H] = \frac{|D_3|}{|H|}= 2$. Therefore $H$ is certainly normal since every subgroup of index $2$ is normal.

If $B$ is one of the symmetries in $D_3$ then the quotient group is given by

$$D_3/H= \{H, B + H\}$$

because two transformations will be in the same coset if and only if their images can be rotated one onto another. The rotations will belong to the coset represented by the identity, and the symmetries will belong to the other coset (since they cannot be rotated onto the original image).

Answer 2

If $F$ is the reflection (say through $x$ axis) then since $H$ is normal of index $2$ there are only two cosets: $H$ itself, and $FH$ meaning the set of left multiples of elements of $H$ by $F.$

For question 2, just check all these in $H$ and in $FH$ are pairwise distinct, then you know all six elements of $D_3$ are used.