Alg7.8

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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\isin \mathrm{ker}\phi\, then \phi(ab)=\phi(a)\phi(b)=1\cdot 1=1\, so ab\isin \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(a a^{-1})=\phi(a)\phi(a^{-1})=1\cdot\phi(a^{-1})\implies \forall a\isin \mathrm{ker}\phi, a^{-1}\isin \mathrm{ker}\phi\,

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

Abstract Algebra

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