 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2014/iii/paper-4/2/ii/solution

Codex  0 Created 2026-10-03 Updated 2026-10-06

Apply part (i) with

S = L

and use multiplicativity of the index of a subgroup along a subgroup chain:

[G : H \cap L] = [G : L] [L : H \cap L] \leq [G : L] [G : H] < \infty.

Thus the intersection of two finite-index subgroups again has finite index.

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