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Let $G$ be a group s.t $G = \langle(12),(345)\rangle \subseteq S_5$ acts on the set $X = \{1,2,3,4,5\}$. I want to find all orbits and stabilizers of $G$.

The point I don't understand is that according to the definition of them, they are defined for each element of a set.

Orbit: $$G\cdot x=\{g\cdot x \ \colon g\in G\}$$

Stabilizer: $$G_x=\{g\in G \ \colon g\cdot x=x\}$$

In this case, $G$ permutes multiple elements of $X$ so what are $x$ in this case?

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    $\begingroup$ $G$ is a subset of $S_5$ (subgroup, actually), so why the point-wise definitions of orbit and stabilizer (i.e. $x=1,2,3,4,5$) makes you worry? $\endgroup$
    – user810157
    Sep 20, 2020 at 20:19
  • $\begingroup$ So for example, what are the possible destinations of $1$ when you act on it via an arbitrary element of $G$? (There could conceivably be as many as five, but for this particular group action there are fewer.) And which elements of the entire subgroup $G$ (which has elements in addition to the generators) leave $1$ fixed? $\endgroup$ Sep 20, 2020 at 20:55

2 Answers 2

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The group G is the cyclic group $Z_6$ containing the permutations {(1 2)(345),(354),(12),(345),(12)(354),e}. The orbit 1 is {1,2} and orbit of 3 is {3,4,5}. The stabilizer of 1 is {e,(345),(354)} and for 3 the stabilizer is {e,(12)}. Similarly the orbits and stabilizers of other elements are written down.

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  • $\begingroup$ I understood how you found. Just to clarify the permutation (354) is (345)^2 right? $\endgroup$
    – TFC
    Sep 20, 2020 at 21:45
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    $\begingroup$ Yes. (345)(345)=(354) $\endgroup$ Sep 21, 2020 at 21:50
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Just a few general deductions by using the Orbit-Stabilizer Theorem. Since $G$ has order $6$, in principle the stabilizers might have order $1$ or $2$ or $3$ or $6$ (Lagrange's Theorem); but, $1$ is ruled out because there can't be any orbit of $6$ elements (being $X$ of size $5$), and $6$ is ruled out because there isn't any element of $X$ that is fixed by all the elements of $G$. So, just stabilizers of order $2$ and $3$ are allowed, and accordingly orbits of size $3$ and $2$, respectively. Moreover, the elements of $G$ which move all the $5$ elements of $X$ (namely the products of disjoint $2$- and $3$-cycles), by definition can't be elements of any stabilizer.

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