Bourbaki

Proposition 5

Tag 0590, permanent. Cite it as [Bourbaki, Tag 0590].

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

Let f:G→G′f : G \to G' be a homomorphism of groups with operators and HH and H′H' normal stable subgroups of GG and G′G' respectively such that f(H)⊂H′f(H) \subset H'. The mapping ff is compatible with the equivalence relations defined by HH and H′H'. Let π:G→G/H\pi : G \to G/H and π′:G′→G′/H′\pi' : G' \to G'/H' be the canonical homomorphisms. The mapping fˉ:G/H→G′/H′\bar{f} : G/H \to G'/H' derived from ff by passing to the quotients is a homomorphism.

If x≡y(modH)x \equiv y \pmod{H}, then x−1y∈Hx^{-1}y \in H, whence

f(x)−1f(y)=f(x−1)f(y)=f(x−1y)∈f(H)⊂H′ f(x)^{-1}f(y) = f(x^{-1})f(y) = f(x^{-1}y) \in f(H) \subset H'
and hence f(x)≡f(y)(modH′)f(x) \equiv f(y) \pmod{H'}. The second assertion follows from the universal property of quotient laws (§1§ 1, no. 6).

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.