Let G and H be two groups with operators, f a homomorphism of G into H and N the kernel of f. (a) Let H′ be a stable subgroup of H. The inverse image G′=f−1(H′) is a stable subgroup of G and G′ is normal in G if H′ is normal in H. Further, N is a normal subgroup of G′. If f is surjective, then H′=f(G′) and f defines an isomorphism of G′/N onto H′ on passing to the quotient. (b) Let G′ be a stable subgroup of G. The image H′=f(G′) is a stable subgroup of H and f−1(H′)=G′N=NG′. In particular, f−1(H′)=G′ if and only if N⊂G′. If f is surjective and G′ is normal in G, then H′ is normal in H.
(a) Let x and y be in G′ and α∈Ω; then f(x)∈H′ and f(y)∈H′, whence f(xy−1)=f(x)f(y)−1∈H′, that is xy−1∈G′; hence G′ is a subgroup of G. Now f(xα)=f(x)α∈H′, whence xα∈G′ and therefore G′ is stable. Suppose H′ is normal in H and let x∈G′,y∈G; then f(x)∈H′ and
f(yxy−1)=f(y)f(x)f(y)−1∈H′ whence
yxy−1∈G′; hence
G′ is normal in
G. For all
n∈N,f(n)=e∈H′, whence
N⊂G′; as
N is normal in
G, it is normal in
G′. Finally, if
f is surjective,
f(f−1(A))=A for every subset A of H, whence
H′=f(G′); the restriction of
f to
G′ is a homomorphism
f′ of
G′ onto
H′ of kernel
N, hence
f′ defines on passing to the quotient an isomorphism of
G′/N onto
H′.
(b) Let a and b be in H′ and α in Ω; there exist x,y in G′ with a=f(x) and b=f(y), whence ab−1=f(xy−1)∈H′, hence H′ is a subgroup of H which is stable, for aα=f(xα)∈H′. Let x∈G; then x∈f−1(H′) if and only if f(x)∈H′=f(G′), that is if and only if there exists y in G′ with f(x)=f(y); the relation f(x)=f(y) is equivalent to the existence of n∈N with x=yn; finally, x∈f−1(H′) is equivalent to x∈G′N=NG′. Clearly the relation G′=G′N is equivalent to G′⊃N. Suppose finally that f is surjective and G′ is normal in G; let a∈H′ and b∈H; there exist x∈G′ and y∈G with a=f(x) and b=f(y), whence bab−1=f(yxy−1)∈f(G′)=H′. Hence H′ is normal in H.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
Nothing.