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SSC Pure Mathematics QUESTION #4104
Question 1
In a group homomorphism \(\varphi:G\to H\), the kernel \(\ker(\varphi)\) is ALWAYS:
  • A subgroup of \(H\) only
  • Any arbitrary subset of \(G\)
  • A normal subgroup of \(G\)✔️
  • Equal to \(G\) if \(\varphi\) is surjective
Correct Answer Explanation
The kernel \(\ker(\varphi)=\{g\in G:\varphi(g)=e_H\}\) is always a normal subgroup of the domain \(G\). Proof: for \(k\in\ker(\varphi)\) and any \(g\in G\), \(\varphi(gkg^{-1})=\varphi(g)e_H\varphi(g)^{-1}=e_H\), so \(gkg^{-1}\in\ker(\varphi)\) — confirming normality.