Third isomorphism theorem for modules

If A ⊆ B ⊆ M then (M/A)/(B/A) ≅ M/B.
Third isomorphism theorem for modules

Third isomorphism theorem (modules): Let MM be an RR-module and let ABMA\subseteq B\subseteq M be . Then there is a natural isomorphism

(M/A)/(B/A)    M/B (M/A)/(B/A) \;\cong\; M/B

of RR-modules, where each quotient is a .

This theorem expresses the compatibility of iterated quotients and is fundamental in organizing “modding out step by step” in module theory.

Proof sketch (optional): Use the natural surjection M/AM/BM/A\to M/B induced from MM/BM\to M/B; its kernel is B/AB/A. Apply the first isomorphism theorem to that map.