Bourbaki

Lemma 1 (Zassenhaus)

Tag 059F, permanent. Cite it as [Bourbaki, Tag 059F].

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

Let H and K be two stable subgroups of a group with operators G and H' and K' normal stable subgroups of H and K respectively; then H'.(H \cap K') is a normal stable subgroup of H'.(H \cap K), K'.(K \cap H') is a normal stable subgroup of K'.(K \cap H) and the quotient groups with operators (H'.(H \cap K))/(H'.(H \cap K')) and (K'.(K \cap H))/(K'.(K \cap H')) are isomorphic.

By Theorem 4, H' \cap K = H' \cap (H \cap K) is a normal stable subgroup of H \cap K; similarly K' \cap H is a normal stable subgroup of K \cap H; hence (no. 6, Corollary 2) (H' \cap K)(K' \cap H) is a normal stable subgroup of H \cap K. By Theorem 4 applied to the group H,

H′.(H′∩K).(K′∩H)=H′.(H∩K′) H'.(H' \cap K).(K' \cap H) = H'.(H \cap K')

is a normal stable subgroup of H'.(H \cap K) and the quotient group

(H′.(H∩K))/(H′.(H∩K)) (H'.(H \cap K))/(H'.(H \cap K))

is isomorphic to

(H∩K)/((H′∩K).(K′∩H)). (H \cap K)/((H' \cap K).(K' \cap H)).

In the last quotient, H and H' on the one hand and K and K' on the other appear symmetrically; permuting them the stated result is obtained.

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

Cited by

Nothing.

Cites

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