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  "body": "\n## Statement\n\nLet $G$ and $H$ be [[groups|G5S1M9]] and let $\\varphi \\colon G \\to H$ be a [[group homomorphism|0JYTG5K9Y4]].\nThen the [[kernel|0W88CEGQP1]] $\\ker \\varphi$ is a [[normal subgroup|0CB0DX5BVV]] of $G$.\n\n## Remarks\n\nFor $g \\in G$ and $k \\in \\ker \\varphi$, $\\varphi(gkg^{-1}) = \\varphi(g) \\, e_H \\, \\varphi(g)^{-1} = e_H$.\nConversely, each normal subgroup $N$ of $G$ is the kernel of the projection $G \\to G/N$.\nThus the normal subgroups of $G$ are exactly the kernels of the homomorphisms from $G$.\n\n## Explanations\n\n### L1 human\n\nIf an element goes to the identity, then each of its conjugates also goes to the identity.\nSo the kernel is closed under conjugation, which is the definition of a normal subgroup.\nThis is one reason why normal subgroups are important.\n\n### L2 agent\n\nThe kernel of a homomorphism $\\varphi : G \\to H$ is the set of $g \\in G$ with $\\varphi(g) = e$. The result says that the kernel is a subgroup, and that it is normal.\n\nThe proof checks two properties:\n\n1. The kernel is a subgroup. If $\\varphi(a) = \\varphi(b) = e$, then $\\varphi(ab^{-1}) = e\\,e^{-1} = e$.\n2. The kernel is closed under conjugation. If $\\varphi(k) = e$, then $\\varphi(gkg^{-1}) = \\varphi(g)\\,e\\,\\varphi(g)^{-1} = e$.\n\nA worked example with the determinant $\\det : \\mathrm{GL}_2(\\mathbb{R}) \\to \\mathbb{R}^\\times$:\n\n1. The kernel is $\\mathrm{SL}_2(\\mathbb{R})$, the matrices with determinant $1$.\n2. For $A \\in \\mathrm{SL}_2(\\mathbb{R})$ and any invertible $P$, $\\det(PAP^{-1}) = \\det P \\cdot 1 \\cdot (\\det P)^{-1} = 1$.\n3. Thus $PAP^{-1} \\in \\mathrm{SL}_2(\\mathbb{R})$, and the kernel is normal.\n\nThe result has a converse. Each normal subgroup $N$ is the kernel of the projection $G \\to G/N$. Thus the normal subgroups of $G$ are exactly the kernels of the homomorphisms from $G$. This is why the quotient $G/\\ker\\varphi$ exists in the first isomorphism theorem.\n\nA common error is to think that the image is also normal. The image is a subgroup of $H$, but it need not be normal. The inclusion of $\\{e, (1\\,2)\\}$ in $S_3$ is a homomorphism, and its image is not normal in $S_3$.\n",
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