Bourbaki

Proposition 1

Tag 058N, permanent. Cite it as [Bourbaki, Tag 058N].

Algebra, chapter I, § 4. Groups and groups with operators

Let G be a group with operators and H a subset of G which is stable under the homotheties of G. The following conditions are equivalent:

(a) H is a stable subgroup of G. (b) H is non-empty and the relations x∈H,y∈H  ⟹  xy∈Hx \in H, y \in H \implies xy \in H and x−1∈Hx^{-1} \in H. (c) H is non-empty and the relations x∈H,y∈H  ⟹  xy−1∈Hx \in H, y \in H \implies xy^{-1} \in H. (d) H is stable under the law on G and the law of composition induced on H by the law of composition on G is a group law.

Clearly (a) implies (b). We show that (b) implies (a). It suffices to show that H contains the identity element of G. As the subset H is non-empty, let x∈Hx \in H. Then x−1∈Hx^{-1} \in H and e=xx−1∈He = xx^{-1} \in H. Clearly (b) implies (c). We show that (c) implies (b). First of all since H is non-empty it contains an element x. Hence xx^{-1} = e is an element of H. For every element x of H, x−1=ex−1x^{-1} = ex^{-1} belongs to H; hence the relations x∈H,y∈Hx \in H, y \in H imply x(y−1)−1=xy∈Hx(y^{-1})^{-1} = xy \in H. Clearly (a) implies (d). We show that (d) implies (a): the canonical injection of H into G is a group homomorphism; hence e∈He \in H and the relation x∈Hx \in H implies x−1∈Hx^{-1} \in H (no. 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.