Bourbaki

Proposition 1

Tag 05B5, permanent. Cite it as [Bourbaki, Tag 05B5].

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

Let M be a monoid operating on a set E and A a subset of E. (a) The stabilizer, strict stabilizer and fixer of A are submonoids of M. (b) Let α\alpha be an invertible element of M; if α\alpha belongs to the strict stabilizer (resp. fixer) of A, so does α−1\alpha^{-1}.

Let e be the identity element of M; then ea=aea = a for every element a∈Aa \in A and therefore e belongs to the fixer of A. Let α\alpha and β\beta be elements of E which stabilize A. Then (αβ)A=α(βA)⊂αA⊂A(\alpha \beta)A = \alpha(\beta A) \subset \alpha A \subset A and therefore the stabilizer of A is a submonoid of M. Similarly for the strict stabilizer and fixer of A, whence (a). If αA=A\alpha A = A, then A=α−1(αA)=α−1AA = \alpha^{-1}(\alpha A) = \alpha^{-1}A. If for all a∈Aa \in A, αa=a\alpha a = a, then a=α−1(αa)=α−1aa = \alpha^{-1}(\alpha a) = \alpha^{-1}a, whence (b).

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

Cited by

Nothing.

Cites

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