Bourbaki

Proposition 2

Tag 05B7, permanent. Cite it as [Bourbaki, Tag 05B7].

Algebra, chapter I, § 5. Groups operating on a set

Let G be a group operating on a set E and, for all x∈Ex \in E, let SxS_x be the stabilizer of x. For all α∈G\alpha \in G, Sαx=αSxα−1S_{\alpha x} = \alpha S_x \alpha^{-1}.

If s∈Sxs \in S_x, then αsα−1(αx)=αsx=αx\alpha s \alpha^{-1}(\alpha x) = \alpha s x = \alpha x, whence αSxα−1⊂Sαx\alpha S_x \alpha^{-1} \subset S_{\alpha x}. As x=α−1(αx)x = \alpha^{-1}(\alpha x), α−1Sαxα⊂Sx\alpha^{-1} S_{\alpha x} \alpha \subset S_x, whence Sαx⊂αSxα−1S_{\alpha x} \subset \alpha S_x \alpha^{-1}.

It is seen similarly that, if A and B are two subsets of E and T is the transporter (resp. strict transporter) of A to B, then the transporter (resp. strict transporter) of αA\alpha A to αB\alpha B is equal to αTα−1\alpha T \alpha^{-1}.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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Cites

Nothing.

Transcribed from the printed edition, 1998, Springer. Not translated.