# 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?

• Is it as simple as observing that $D_8 \circ s =S$ which holds $\forall s \in S$? so also for our specific element. – Wesley Strik Mar 11 at 0:54
• And then we simply construct all elements by composition with the 8 elements of $D_8$, since $S$ consists of a single orbit anyway, it does not matter which elements we have, we can construct the entire orbit. – Wesley Strik Mar 11 at 1:16

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 $$S= \{ (1 \ 2 ) ,(1 \ 4 \ 2 \ 3 ) ,(2 \ 4 \ 3) ,(1 \ 3 \ 4) , (1 \ 3 \ 2 \ 4 ) ,(1 \ 4 \ 2 ) , (3 \ 4 ) , (1 \ 2 \ 3 ) \}$$