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maybe that's idiot, but I'm missing something here. Let $X = \{(123),(132),(124),(142),(134),(143),(234),(243) \}$, $A_4$ act on $X$ by conjugation (inner automorphisms) and $x = (123)$, then $4 =|\mathscr{O}(x)| = |G|/|G_x| = 12/|G_x|$. However, $G_x = \{ 1 \}$ !!!!What's wrong here?

Thanks in advance.

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The stabilizer is not the identity, because your action in conjugation, not multiplication. Every group element stabilizes itself under conjugation, and every power of a group element stabilizes the original element, too.

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  • $\begingroup$ I agree, but this leads to another problem. $|G_x| = 5$ now. $\endgroup$ – user40276 Apr 24 '13 at 17:14
  • $\begingroup$ Check again, you should find three elements in $G_x = \{ g \in A_4 : g x g^{-1} = x \}$. In the case of the conjugation action, the stabilizer of $x$ is often called the center of $x$. $\endgroup$ – Michael Joyce Apr 24 '13 at 17:47
  • $\begingroup$ @MichaelJoyce But the centralizer of $x$ will be composed by the powers of $(123)$, i.e, $(123)^k$ for $k = -2, -1, 0, 1, 2$. What's wrong? Am I missing something. $\endgroup$ – user40276 Apr 24 '13 at 18:02
  • $\begingroup$ $(123)^{-1} = (123)^2$ and $(123)^{-2} = (123)$ :) $\endgroup$ – Michael Joyce Apr 24 '13 at 19:37
  • $\begingroup$ Now, I'm feeling a complete idiot. Thanks $\endgroup$ – user40276 Apr 25 '13 at 2:37

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