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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?

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  • $\begingroup$ Is it as simple as observing that $D_8 \circ s =S$ which holds $\forall s \in S$? so also for our specific element. $\endgroup$ – Wesley Strik Mar 11 at 0:54
  • $\begingroup$ 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. $\endgroup$ – Wesley Strik Mar 11 at 1:16
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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 $$ S= \{ (1 \ 2 ) ,(1 \ 4 \ 2 \ 3 ) ,(2 \ 4 \ 3) ,(1 \ 3 \ 4) , (1 \ 3 \ 2 \ 4 ) ,(1 \ 4 \ 2 ) , (3 \ 4 ) , (1 \ 2 \ 3 ) \}$$

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