Compute a set $S$ given information about how it is acted upon transitively by $D_8$ 
Let $D_8=D_{2 \cdot 4}$ be the dihedral group on a regular $4$-gon. Suppose that  $S$ is a subset of $S_4$, such that S contains the element $( 1 \ 2 \ 3)$. We also know that $D_8$ acts transitively on $S$ via the action:
  $$ D_8 \times S \mapsto S$$
$$ (\sigma, s) \mapsto  \sigma \circ s $$
  Task: determine $S$


My approach:

There are a lot of details, but we have to start somewhere. We can start with the orbit-stabiliser theorem:
$$ |D_8|=|\text{Stab}(( 1 \ 2 \ 3))|\cdot |\text{Orb}(( 1 \ 2 \ 3))|$$
Because we know that $D_8$ acts transivitely on our set $S$, the orbits is all of $S$, 
$$ 8=|\text{Stab}(( 1 \ 2 \ 3))|\cdot |S|$$
We know there is only one element in $D_8$ that fixes/stabilises $(1 \ 2 \ 3)$, this is the identity symmetry, $Id$ and consequently  $|\text{Stab}(( 1 \ 2 \ 3))|=1$. This is because the non-identity symmetries of a square are reflections (these leave two corners fixed) and rotations (the centre is fixed, but none of the corners), neither of these have three fixed non-collinear points, the only symmetry that does this is the identity, therefore it lets the cycle $( 1 \ 2 \ 3)$ do its job, and cycle around three elements, without changing the order. Therefore we have:
$$ |S|=8$$
How do I now determine which $8-1=7$ other elements we are dealing with?
 A: We know that $D_8$ acts transitively on $S$, which means that the action possesses only a single group orbit, which is just the set $S$ itself, i.e., for every $s \in S$, it holds that $D_8 \circ s = S$. We know that $(1 \ 2 \ 3)\in S$, we can use this element to construct the entire orbit, which is simply the set $S$ itself. $S$. We have $D_8=(\langle 
(1 \ 3), (1 \ 2 \ 3 \ 4)\rangle)$ which consists of all symmetries of the square, a regular 4-gon. It contains 4 reflections and 4 rotations (amongst which the identity rotation). We will list the composition of $(1 \  2 \ 3 )$ with all these elements:
$(1 \ 3) \circ (1 \ 2 \ 3)= (1 \ 2 )$
$(2 \ 4) \circ (1 \ 2 \ 3)= (1 \ 4 \ 2 \ 3 )$
$(1 \ 2) (3 \ 4) \circ (1 \ 2 \ 3)=(2 \ 4 \ 3) $
$(1 \ 4) (2 \ 3) \circ (1 \ 2 \ 3)=(1 \ 3 \ 4) $
$(1 \ 2\ 3 \ 4) \circ (1 \ 2 \ 3)= (1 \ 3 \ 2 \ 4 ) $
$(1 \ 3) (2 \ 4) \circ (1 \ 2 \ 3)=(1 \ 4 \ 2 ) $
$(1 \ 4 \ 3 \ 2 )\circ (1 \ 2 \ 3)= (3 \ 4 ) $
$ Id \circ (1 \ 2 \ 3)= (1 \ 2 \ 3 )$
Consequently  $$ S= \{  
   (1 \ 2 )
,(1 \ 4 \ 2 \ 3 )
,(2 \ 4 \ 3) 
,(1 \ 3 \ 4) 
   , (1 \ 3 \ 2 \ 4 ) 
 ,(1 \ 4 \ 2 ) 
  , (3 \ 4 ) 
   , (1 \ 2 \ 3 )  
   \}$$
