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text: 'Given a [-3gd] $G$ and a subgroup $H$, the [4j4 left cosets] of $H$ in $G$ [set_partition partition] $G$, in the sense that every element of $g$ is in precisely one coset.\n\n# Proof\nFirstly, every element is in a coset: since $g \\in gH$ for any $g$.\nSo we must show that no element is in more than one coset.\n\nSuppose $c$ is in both $aH$ and $bH$.\nThen we claim that $aH = cH = bH$, so in fact the two cosets $aH$ and $bH$ were the same.\nIndeed, $c \\in aH$, so there is $k \\in H$ such that $c = ak$.\nTherefore $cH = \\{ ch : h \\in H \\} = \\{ akh : h \\in H \\}$.\n\nExercise: $\\{ akh : h \\in H \\} = \\{ ar : r \\in H \\}$.\n%%hidden(Show solution):\nSuppose $akh$ is in the left-hand side.\nThen it is in the right-hand side immediately: letting $r=kh$.\n\nConversely, suppose $ar$ is in the right-hand side.\nThen we may write $r = k k^{-1} r$, so $a k k^{-1} r$ is in the right-hand side; but then $k^{-1} r$ is in $H$ so this is exactly an object which lies in the left-hand side.\n%%\n\nBut that is just $aH$.\n\nBy repeating the reasoning with $a$ and $b$ interchanged, we have $cH = bH$; this completes the proof.\n\n# Why is this interesting?\n\nThe fact that the left cosets partition the group means that we can, in some sense, "compress" the group $G$ with respect to $H$.\nIf we are only interested in $G$ "up to" $H$, we can deal with the partition rather than the individual elements, throwing away the information we're not interested in.\n\nThis concept is most importantly used in defining the [-4tq].\nTo do this, the subgroup must be [4h6 normal] ([4h9 proof]).\nIn this case, the collection of cosets itself inherits a group structure from the parent group $G$, and the structure of the quotient group can often tell us a lot about the parent group.',
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