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CSS Pure Mathematics QUESTION #1278
Question 1
If a homomorphism \(\varphi: G\to H\) is injective, then:
  • Every element of H has a preimage in G
  • \(\ker\varphi\) is trivial (contains only identity)✔️
  • G and H have the same order
  • \(\varphi\) must also be surjective
Correct Answer Explanation
An injective (one-to-one) homomorphism has trivial kernel: if \(\varphi(g)=e_H\) then \(\varphi(g)=\varphi(e_G)\), and by injectivity \(g=e_G\). So \(\ker\varphi=\{e_G\}\). This is both a necessary and sufficient condition for injectivity of a group homomorphism.