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Suppose a group $G$ of order 10 acts on an infinite set $X$

Q1: What are all the possible sizes of the orbits of $G$?

Q2: How many orbits are there?

Now suppose $X$ has 8 elements. What are the possible sizes of orbits of $G$?

Now suppose $X$ has 11 elements. What’s the minimal number of $G$ orbits in $X$?

My attempt:

There could be some fixed point $x \in X$, but there could also exist a point that is moved to different points upon the action of different $g \in G$, so possible sizes of orbits would range from 1 to 10. It can also be the case that all points are fixed points under this action such that the number of orbits = cardinality of X or every $x \in X$ can be obtained by the action of some $g$ on a $y \in X$ such that there is only one orbit.

For $X$ having 8 elements, orbits can have sizes 1 to 8

For $X$ having 11 elements, there have to be at least two orbits since maximum size of one orbit is 10.

Is this correct?

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Q1. By the orbit-stabiliser theorem, orbits can only have lengths 1,2,5 or 10.

Q2. Therefore there have to be infinitely many orbits.

For $|X|=8$, orbits can only have lengths 1,2 or 5.

You are correct for the $|X|=11$ case.

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  • $\begingroup$ G/G_x is isomorphic to G.x, which is the orbit of x. Knowing that cardinality of G is 10, G_x would have to be a divisor for the theorem to make sense. So, we know orbits can only have lengths 1,2,5 or 10. Is this the right reasoning? $\endgroup$ Oct 8 '19 at 20:29
  • $\begingroup$ Yes, that's right. $\endgroup$
    – S. Dolan
    Oct 8 '19 at 20:35

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