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Prove that the kernel of a homomorphism \phi :G\to G'\, is a subgroup of G\,. In this proof the identity element is represented by 1\,.

Closure: If two elements a,b\in {\mathrm  {ker}}\phi \, then \phi (ab)=\phi (a)\phi (b)=1\cdot 1=1\, so ab\in {\mathrm  {ker}}\phi \,.

Identity: 1=1\cdot 1\implies \phi (1)=\phi (1\cdot 1)=\phi (1)\phi (1)\,.

Apply \phi (1)^{{-1}}\, to both sides to obtain 1=\phi (1)\,.

Inverse: 1=\phi (1)=\phi (aa^{{-1}})=\phi (a)\phi (a^{{-1}})=1\cdot \phi (a^{{-1}})\implies \forall a\in {\mathrm  {ker}}\phi ,a^{{-1}}\in {\mathrm  {ker}}\phi \,

Associativity: If \exists a,b,c\in {\mathrm  {ker}}\phi \mid (ab)c\neq a(bc)\, then \exists a,b,c\in G\mid (ab)c\neq a(bc)\, which would contradict the fact that G\, is a group.

Abstract Algebra

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