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Consider the action of the full symmetric group $S_3$ on the cube $[0,2] \times [0,2] \times [0,2]$.

Classify the orbits of this action and determine their cardinalities.

My Answer: What I note is that the orbit can have six possibilities: $(j,i,k)$, $(k,i,j)$, $(i,j,k)$, $(j,k,i)$, $(k,i,j)$; of the form $(i,i,k)$ with stabilizer $2$ and orbit $3$; or of the form $(i,i,i)$ just a stabilizer.

So does this mean that the orbits of the action we're looking at can be $(0,2,0)$, $(0,0,2)$, $(2,0,0)$; and $(2,2,0)$ or $(0,2,2)$ or $(2,0,2)$?

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Hint: You correctly determined the orbits of the elements $(0,2,0)$ and $(2,2,0)$. But notice that the interval $[0,2]$ has more elements than just $0$ and $2$. Therefore, most orbits will have six elements. For instance, consider the orbit of the element $(0,1,2)$. It consists of the elements $ (0,1,2), (0,2,1), (1,0,2), (1,2,0), (2,0,1), (2,1,0). $ Finally, the orbit of the element $(1,1,1)$ consists of only one element.

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How does $S_3$ act on the cube? By permutation of coordinates? Also, your terminology is slightly difficult to follow, even though I believe you mean the right thing. Your last line doesn't quite make sense, though. Why are you only looking at elements with entries 0 and 2?

But in principle you are right. If we are talking about the same action, then there are three different cases:

  1. Orbits of elements with all three coordinates the same, these orbits are singletons.

  2. Orbits of elements with exactly two coordinates the same, these are of length 3.

  3. Orbits of elements with all three coordinates different. These are of length 6.

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    Can you list out the orbits. You did clearly say the cardinalities, but please list them out i..e, the elements which give rise to how the cardinalities came about2012-01-25