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text: '[summary: Lagrange's Theorem gives us a restriction on the size of a subgroup of a finite group: namely, subgroups have size dividing the parent group.]\n\nGiven a finite [-3gd] $G$, it may have many [576 subgroups].\nSo far, we know almost nothing about those subgroups; it would be great if we had some way of restricting them.\n\nAn example of such a restriction, which we do already know, is that a subgroup $H$ of $G$ has to have [3gg size] less than or equal to the size of $G$ itself.\nThis is because $H$ is contained in $G$, and if the set $X$ is contained in the set $Y$ then the size of $X$ is less than or equal to the size of $Y$.\n(This would have to be true for any reasonable definition of "size"; the [4w5 usual definition] certainly has this property.)\n\nLagrange's Theorem gives us a much more powerful restriction: not only is the size $|H|$ of $H$ less than or equal to $|G|$, but in fact $|H|$ divides $|G|$.\n\n%%hidden(Example: subgroups of the cyclic group on six elements):\n*A priori*, all we know about the subgroups of the [-47y] $C_6$ of order $6$ is that they are of order $1, 2, 3, 4, 5$ or $6$.\n\nLagrange's Theorem tells us that they can only be of order $1, 2, 3$ or $6$: there are no subgroups of order $4$ or $5$.\nLagrange tells us nothing about whether there *are* subgroups of size $1,2,3$ or $6$: only that if we are given a subgroup, then it is of one of those sizes.\n\nIn fact, as an aside, there are indeed subgroups of sizes $1,2,3,6$:\n\n- the subgroup containing only the identity is of order $1$\n- the "improper" subgroup $C_6$ is of order $6$\n- subgroups of size $2$ and $3$ are guaranteed by [4l6 Cauchy's theorem].\n\n%%\n\n# Proof\n\nIn order to show that $|H|$ divides $|G|$, we would be done if we could divide the elements of $G$ up into separate buckets of size $|H|$.\n\nThere is a fairly obvious place to start: we already have one bucket of size $|H|$, namely $H$ itself (which consists of some elements of $G$).\nCan we perhaps use this to create more buckets of size $|H|$?\n\nFor motivation: if we think of $H$ as being a collection of symmetries (which we can do, by [49b Cayley's Theorem] which states that all groups may be viewed as collections of symmetries), then we can create more symmetries by "tacking on elements of $G$".\n\nFormally, let $g$ be an element of $G$, and consider $gH = \\{ g h : h \\in H \\}$.\n\nExercise: every element of $G$ does have one of these buckets $gH$ in which it lies.\n%%hidden(Show solution):\nThe element $g$ of $G$ is contained in the bucket $gH$, because the identity $e$ is contained in $H$ and so $ge$ is in $gH$; but $ge = g$.\n%%\n\nExercise: $gH$ is a set of size $|H|$. %%note:More formally put, [-4j8].%%\n%%hidden(Show solution):\nIn order to show that $gH$ has size $|H|$, it is enough to match up the elements of $gH$ [499 bijectively] with the elements of $|H|$.\n\nWe can do this with the [-3jy] $H \\to gH$ taking $h \\in H$ and producing $gh$.\nThis has an [4sn inverse]: the function $gH \\to H$ which is given by pre-multiplying by $g^{-1}$, so that $gx \\mapsto g^{-1} g x = x$.\n%%\n\nNow, are all these buckets separate? Do any of them overlap?\n\nExercise: if $x \\in rH$ and $x \\in sH$ then $rH = sH$. That is, if any two buckets intersect then they are the same bucket. %%note:More formally put, [4j5].%%\n%%hidden(Show solution):\nSuppose $x \\in rH$ and $x \\in sH$.\n\nThen $x = r h_1$ and $x = s h_2$, some $h_1, h_2 \\in H$.\n\nThat is, $r h_1 = s h_2$, so $s^{-1} r h_1 = h_2$.\nSo $s^{-1} r = h_2 h_1^{-1}$, so $s^{-1} r$ is in $H$ by closure of $H$.\n\nBy taking inverses, $r^{-1} s$ is in $H$.\n\nBut that means $\\{ s h : h \\in H \\}$ and $\\{ r h : h \\in H\\}$ are equal.\nIndeed, we show that each is contained in the other.\n\n- if $a$ is in the right-hand side, then $a = rh$ for some $h$. Then $s^{-1} a = s^{-1} r h$; but $s^{-1} r$ is in $H$, so $s^{-1} r h$ is in $H$, and so $s^{-1} a$ is in $H$.\nTherefore $a \\in s H$, so $a$ is in the left-hand side.\n- if $a$ is in the left-hand side, then $a = sh$ for some $h$. Then $r^{-1} a = r^{-1} s h$; but $r^{-1} s$ is in $H$, so $r^{-1} s h$ is in $H$, and so $r^{-1} a$ is in $H$.\nTherefore $a \\in rH$, so $a$ is in the right-hand side.\n%%\n\nWe have shown that the "[4j4 cosets]" $gH$ are all completely disjoint and are all the same size, and that every element lies in a bucket; this completes the proof.',
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