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Let $X=\{\{1,2\},\{1,3\},\{1,4\},\{2,3\},\{2,4\},\{3,4\}\}$ and $G=S_4$

(a) Show that $\phi \{x_1,x_2\}=\{\phi (x_1),\phi (x_2)\}$ determines an action of $S_4$ on $X$, where $\phi \in G$

(b) Determine the number of orbits of this action.

(a) I’m not really sure how to explain this but it comes as an intuitive action that if you act $\phi$ on $x_1$, it is just mapping $x_1$ to $\phi (x_1)$, which essentially is just mapping 1 to $\phi (1)$. But I’m not sure how to present this formally.

(b) Any hints on how I can do this? I’m not sure if I should write out all 24 elements in $S_4$ and use Burnside’s Theorem to find the number of orbits. But if I do so it might be too tedious. What happens when the it becomes larger? i.e $S_5$ acting on $X$; I don’t think I can write out all 120 elements.

Wondering if anyone can help with this?

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  • $\begingroup$ What are $x_1$ and $x_2$ denoting? I guess they range over $\{1,2,3,4\}$. $\endgroup$ – Berci Nov 27 '18 at 11:10
  • $\begingroup$ @Berci Yes. OP is acting on a graph. $\endgroup$ – Pedro Tamaroff Nov 27 '18 at 11:16
  • $\begingroup$ If we take $g=(1,2) \in S_4$ then $g$ reverses the co-ordinates of each member of $X$ (I am assuming that the members of $X$ are ordered tuples rather than unordered sets). But $\{2,1\}, \{3,1\}$ etc, are not in $X$. So I am puzzled as to how the operation specified is an action on $X$ if $X$ only consists of the six points listed. Am I missing something ? $\endgroup$ – gandalf61 Nov 27 '18 at 11:20
  • $\begingroup$ @gandalf61 Yes it’s supposed to be an unordered set $\endgroup$ – Icycarus Nov 27 '18 at 11:34
  • $\begingroup$ @Icycarus But in that case each member of $G$ simply maps each member of $X$ to itself, so this is a very uninteresting action ! Is it possible that the operation that is intended is $\phi\{x_1,x_2\} = \{\phi(x_1), \phi(x_2)\}$ ? $\endgroup$ – gandalf61 Nov 27 '18 at 11:51
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a) Note that $X$ contains all 2 elements subsets of $\{1,2,3,4\}$, so $X$ is indeed closed under the given map and that map satisfies the condition of being a group action.

b) Start with an element of $X$ and find its orbit. If there's any element left off, find its orbit too, and so on..

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a) To formally show that $\phi\{x_1,x_2\}=\{\phi(x_1),\phi(x_2)\}$ is an action, you need to show two things:

1) $e\{x_1,x_2\}=\{x_1,x_2\} \space \forall \{x_1,x_2\} \in X$ where $e$ is the identity in $G$.

2) $\phi_1(\phi_2\{x_1,x_2\}) = (\phi_1 \phi_2)\{x_1,x_2\} \space \forall \{x_1,x_2\} \in X \space \forall \phi_1, \phi_2 \in G$

These are both straightforward.

b) Given two elements of $X$, think about how you can find a member of $G$ that maps one to the other (in fact, there is more than one as user25959 hints at above). This shows that $G$ acts transitively on $X$, so there is only one orbit.

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  • $\begingroup$ i) the identity element is (1)(2)(3)(4) so applying $e\{x_1,x_2\}=\{x_1,x_2\}$ so it’s straightforward. How about (ii)? How do we go about showing it? $\endgroup$ – Icycarus Nov 27 '18 at 20:17
  • $\begingroup$ @Icycarus For (ii) here is an example. To find a permutation in $g \in G$ that maps $\{1.2\}$ to $\{3,4\}$, you could start by requiring $g$ to map $1$ to $3$ and $2$ to $4$. Now $3$ can either map to $1$, which gives $g=(13)(24)$, or $3$ can map to $2$ which gives $g=(1324)$. Note that $3$ cannot map to $4$ because $2$ already maps to $4$. Alternatively, you can require $g$ to map $1$ to $4$ and $2$ to $3$, which gives $g=(14)(23)$ or $g=(1423)$. So there are four possibilities for $g$. $\endgroup$ – gandalf61 Nov 28 '18 at 8:54
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The stabilizer of $\{i,j\}$ consists of the permutations $e, (ij), (kl)$, and $(ij)(kl)$ where $k,l$ are the "other two" elements of $\{1,2,3,4\}$. So by the orbit-stabilizer theorem, $|Orb_{\{i,j\}}|\cdot 4 = |S_4|= 24$.

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