Bourbaki

Theorem 3

Tag 0595, permanent. Cite it as [Bourbaki, Tag 0595].

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

Let f:G→G′f : G \to G' be a homomorphism of groups with operators. (a) Ker(f)\mathrm{Ker}(f) is a normal stable subgroup of GG; (b) Im(f)\mathrm{Im}(f) is a stable subgroup of G′G'; (c) the mapping ff is compatible with the equivalence relation defined on GG by Ker(f)\mathrm{Ker}(f); (d) the mapping f~:G/Ker(f)→Im(f)\tilde{f} : G/\mathrm{Ker}(f) \to \mathrm{Im}(f) derived from ff by passing to the quotient is an isomorphism of groups with operators; (e) f=i∘f~∘πf = i \circ \tilde{f} \circ \pi, where ii is the canonical injection of Im(f)\mathrm{Im}(f) into G′G' and π\pi is the canonical homomorphism of GG onto G/Ker(f)G/\mathrm{Ker}(f).

Assertion (b) follows from Proposition 6. The equivalence relation f(x)=f(y)f(x) = f(y) on GG is compatible with the group with operators structure on GG. By Theorem 2 (no. 4), it is therefore of the form y∈xHy \in xH, where HH is a normal stable subgroup of GG and HH is the class of the identity element, whence H=Ker(f)H = \mathrm{Ker}(f). Assertions (a), (c) and (d) then follow. Assertion (e) is obvious (Set Theory, II, § 6, no. 5).

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

Cited by

Cites

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