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  title: 'Index two subgroup of group is normal',
  clickbait: 'An easy (though not very widely applicable) criterion for a subgroup to be normal.',
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  text: 'Let $H$ be a [-subgroup] of the [-3gd] $G$, of [index_of_subgroup index] $2$.\nThen $H$ is a [-4h6] of $G$.\n\n# Proof\n\nWe must show that $H$ is closed under [4gk conjugation] by elements of $G$.\n\nSince $H$ has index $2$ in $G$, there are two left [group_coset cosets]: $H$ and $xH$ for some specific $x$.\nThere are also two right cosets: $H$ and $Hy$.\n\nNow, since $x \\not \\in H$, it must be the case that $x \\in Hy$; so without loss of generality, $x = y$.\n\nHence $xH = Hx$ and so $xHx^{-1} = H$.\n\nIt remains to show that $H$ is closed under conjugation by *every* element of $G$.\nBut every element of $G$ is either in $H$, or in $xH$; so it is either $h$ or $xh$, for some $h \\in H$.\n\n- $hHh^{-1}$ is equal to $H$ since $hH = H$ and $Hh^{-1} = H$.\n- $xh H (xh)^{-1} = xhHh^{-1} x^{-1} = xHx^{-1} = H$.\n\nThis completes the proof.\n',
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