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  title: 'Subgroup is normal if and only if it is the kernel of a homomorphism',
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  text: 'Let $N$ be a [-subgroup] of [-3gd] $G$.\nThen $N$ is [4h6 normal] in $G$ if and only if there is a group $H$ and a [-47t] $\\phi:G \\to H$ such that the [49y kernel] of $\\phi$ is $N$.\n\n# Proof\n\n## "Normal" implies "is a kernel"\nLet $N$ be normal, so it is closed under [4gk conjugation].\nThen we may define the [-quotient_group] $G/N$, whose elements are the [group_coset left cosets] of $N$ in $G$, and where the operation is that $gN + hN = (g+h)N$.\nThis group is well-defined ([quotient_by_subgroup_is_well_defined_iff_normal proof]).\n\nNow there is a homomorphism $\\phi: G \\to G/N$ given by $g \\mapsto gN$.\nThis is indeed a homomorphism, by definition of the group operation $gN + hN = (g+h)N$.\n\nThe kernel of this homomorphism is precisely $\\{ g : gN = N \\}$; that is simply $N$:\n\n- Certainly $N \\subseteq \\{ g : gN = N \\}$ (because $nN = N$ for all $n$, since $N$ is closed as a subgroup of $G$);\n- We have $\\{ g : gN = N \\} \\subseteq N$ because if $gN = N$ then in particular $g e \\in N$ (where $e$ is the group identity) so $g \\in N$.\n\n## "Is a kernel" implies "normal"\nLet $\\phi: G \\to H$ have kernel $N$, so $\\phi(n) = e$ if and only if $n \\in N$.\nWe claim that $N$ is closed under conjugation by members of $G$.\n\nIndeed, $\\phi(h n h^{-1}) = \\phi(h) \\phi(n) \\phi(h^{-1}) = \\phi(h) \\phi(h^{-1})$ since $\\phi(n) = e$.\nBut that is $\\phi(h h^{-1}) = \\phi(e)$, so $hnh^{-1} \\in N$.\n\nThat is, if $n \\in N$ then $hnh^{-1} \\in N$, so $N$ is normal.',
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