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    The kernel of a homomorphism is a normal subgroup

    P-JX3FO

    Area
    Algebra
    Topics
    Homomorphisms, Normal Subgroups
    Status
    draft
    • Prove that the kernel of a homomorphism is a normal subgroup.
    Solution

    Let \(\phi:G\to H\) be a group homomorphism. Its kernel is \[ \ker\phi=\{g\in G:\phi(g)=1_H\}. \] It is a subgroup: if \(x,y\in\ker\phi\), then \[ \phi(xy^{-1})=\phi(x)\phi(y)^{-1}=1_H. \]

    For normality, let \(g\in G\) and \(k\in\ker\phi\). Then \[ \phi(gkg^{-1})=\phi(g)\phi(k)\phi(g)^{-1}=1_H, \] so \(gkg^{-1}\in\ker\phi\). Therefore \[ \ker\phi\trianglelefteq G. \]

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