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I'm got some difficulties with something. We consider G a group of order 24, which not contain any element of order 6, and we consider $\sigma \in G$ an element of order 3 in G. We consider as well the conjugation action : $G \times G \rightarrow G$ defined by $(g,\sigma) \rightarrow g\sigma g^{-1}$, and I want to show that the stabilizer of $\sigma$ for this action is $\left<\sigma\right>$. The inclusion $\left<\sigma\right> \subseteq Stab(\sigma)$ is done, but I don't succeed to demonstrate the other inclusion.

Someone have an idea and could help me ? :)

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In effect you are looking at the centraliser $C=C_G(\sigma)$ which certainly contains $\left<\sigma\right>$. If its order was $>3$ its order would be $6$, $12$ or $24$ and be even. So $C$ would contain an element $\tau$ of order $2$. Then $\sigma\tau$ would have order $6$.

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  • $\begingroup$ Oh yes, thank you very much ! $\endgroup$ – ChocoSavour Dec 2 '17 at 11:24

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