Bourbaki

Proposition 19

Tag 05A1, permanent. Cite it as [Bourbaki, Tag 05A1].

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

*Let a be an integer > 0. Let H be a subgroup of Z/aZ\mathbf{Z}/a\mathbf{Z}, b the order of H and c its index in Z/aZ\mathbf{Z}/a\mathbf{Z}. Then a=bca = bc, H=cZ/aZH = c\mathbf{Z}/a\mathbf{Z} and H is isomorphic to Z/bZ\mathbf{Z}/b\mathbf{Z}.

Conversely, let b and c be two integers > 0 such that a=bca = bc. Then aZ⊂cZa\mathbf{Z} \subset c\mathbf{Z} and cZ/aZc\mathbf{Z}/a\mathbf{Z} is a subgroup of Z/aZ\mathbf{Z}/a\mathbf{Z}, of order b and index c.

a=bca = bc by no. 4, Corollary to Proposition 4. By no. 7, Theorem 4, H is of the form H′/aZH'/a\mathbf{Z}, where H′H' is a subgroup of Z\mathbf{Z} and Z/H′\mathbf{Z}/H' is isomorphic to (Z/aZ)/H(\mathbf{Z}/a\mathbf{Z})/H and hence of order c. By Proposition 17 and the Corollary to Proposition 16, H′=cZH' = c\mathbf{Z} and hence H is monogenous. Finally, H is isomorphic to Z/bZ\mathbf{Z}/b\mathbf{Z} by Proposition 18. Conversely, if a=bca = bc, then aZ⊂cZa\mathbf{Z} \subset c\mathbf{Z} for a∈cZa \in c\mathbf{Z}: the quotient group (Z/aZ)/(cZ/aZ)(\mathbf{Z}/a\mathbf{Z})/(c\mathbf{Z}/a\mathbf{Z}) is isomorphic to Z/cZ\mathbf{Z}/c\mathbf{Z} (no. 7, Theorem 4) and hence of order c (no. 4, Corollary to Proposition 4) and index b (no. 4, Corollary to Proposition 4).

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

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