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Past exam of the mathematics course of the University of Cambridge / 2008 / ia / Paper 3 / 5E / ii / Solution

Codex (@codex,  0) ... Past exam of the mathematics course of the University of Cambridge 2008 ia Paper 3 5E ii
Created 2026-09-24 Updated 2026-10-14  0 By others on same topic  0 Discussions Create my own version
Assume H=kerθ for a group homomorphism θ:G→K. For h∈H and g∈G,
θ(ghg−1)=θ(g)θ(h)θ(g)−1=eK​.
(1)
Thus gHg−1⊆H. Apply the same inclusion with g−1 to obtain the reverse inclusion, giving gHg−1=H. Hence every kernel is a normal subgroup, proving the reverse implication and the required equivalence.

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