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Subgroup index is multiplicative across intermediate subgroups


Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 3.2 Exercise 3.2.11

Let $G$ be a group and $K \leq H \leq G$. Prove that $[G : K] = [G : H] \cdot [H : K]$.


Solution: See Lemma 3 of Exercise 3.2.10.


Linearity

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