{
  "index": "1969-B-2",
  "type": "NT",
  "tag": [
    "NT",
    "ALG"
  ],
  "difficulty": "",
  "question": "B-2. Show that a finite group can not be the union of two of its proper subgroups. Does the statement remain true if \"two\" is replaced by \"three\"?",
  "solution": "B-2 The number of elements in a subgroup is a divisor of the order of the group. Thus a proper subgroup can have no more than half of all the elements. Two subgroups always have the identity in common and hence their union cannot be the entire group.\n\nAn example for the second part of the problem is the Klein group, which has an identity and three elements \\( x, y, z \\) of order two. The product of any two distinct elements from \\( \\{x, y, z\\} \\) is the third. This group is the union of three proper subgroups.\n\nAlternate Solution: Let \\( G=H \\cup K \\), with \\( H \\) and \\( K \\) proper subgroups. There exists \\( k \\in K \\) with \\( k \\notin H \\). None of the elements in \\( k H \\) are in \\( H \\) and so \\( k H \\subset K \\). Hence \\( H \\subset k^{-1} K=K \\) and \\( K=H \\cup K=G \\), a contradiction.",
  "vars": [
    "G",
    "H",
    "K",
    "k",
    "x",
    "y",
    "z"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "G": "groupall",
        "H": "subgrpone",
        "K": "subgrptwo",
        "k": "elementk",
        "x": "elemfirst",
        "y": "elemsecond",
        "z": "elemthird"
      },
      "question": "B-2. Show that a finite group can not be the union of two of its proper subgroups. Does the statement remain true if \"two\" is replaced by \"three\"?",
      "solution": "B-2 The number of elements in a subgroup is a divisor of the order of the group. Thus a proper subgroup can have no more than half of all the elements. Two subgroups always have the identity in common and hence their union cannot be the entire group.\n\nAn example for the second part of the problem is the Klein group, which has an identity and three elements \\( elemfirst, elemsecond, elemthird \\) of order two. The product of any two distinct elements from \\( \\{elemfirst, elemsecond, elemthird\\} \\) is the third. This group is the union of three proper subgroups.\n\nAlternate Solution: Let \\( groupall=subgrpone \\cup subgrptwo \\), with \\( subgrpone \\) and \\( subgrptwo \\) proper subgroups. There exists \\( elementk \\in subgrptwo \\) with \\( elementk \\notin subgrpone \\). None of the elements in \\( elementk\\, subgrpone \\) are in \\( subgrpone \\) and so \\( elementk\\, subgrpone \\subset subgrptwo \\). Hence \\( subgrpone \\subset elementk^{-1} subgrptwo=subgrptwo \\) and \\( subgrptwo=subgrpone \\cup subgrptwo=groupall \\), a contradiction."
    },
    "descriptive_long_confusing": {
      "map": {
        "G": "sunflower",
        "H": "cupboard",
        "K": "elephant",
        "k": "teaspoon",
        "x": "glaciers",
        "y": "volcanic",
        "z": "crescent"
      },
      "question": "B-2. Show that a finite group can not be the union of two of its proper subgroups. Does the statement remain true if \"two\" is replaced by \"three\"?",
      "solution": "B-2 The number of elements in a subgroup is a divisor of the order of the group. Thus a proper subgroup can have no more than half of all the elements. Two subgroups always have the identity in common and hence their union cannot be the entire group.\n\nAn example for the second part of the problem is the Klein group, which has an identity and three elements \\( glaciers, volcanic, crescent \\) of order two. The product of any two distinct elements from \\( \\{glaciers, volcanic, crescent\\} \\) is the third. This group is the union of three proper subgroups.\n\nAlternate Solution: Let \\( sunflower=cupboard \\cup elephant \\), with \\( cupboard \\) and \\( elephant \\) proper subgroups. There exists \\( teaspoon \\in elephant \\) with \\( teaspoon \\notin cupboard \\). None of the elements in \\( teaspoon cupboard \\) are in \\( cupboard \\) and so \\( teaspoon cupboard \\subset elephant \\). Hence \\( cupboard \\subset teaspoon^{-1} elephant=elephant \\) and \\( elephant=cupboard \\cup elephant=sunflower \\), a contradiction."
    },
    "descriptive_long_misleading": {
      "map": {
        "G": "isolateelement",
        "H": "supersetwork",
        "K": "outsiderset",
        "k": "aggregation",
        "x": "continuumval",
        "y": "divergentval",
        "z": "infiniteval"
      },
      "question": "B-2. Show that a finite group can not be the union of two of its proper subgroups. Does the statement remain true if \"two\" is replaced by \"three\"?",
      "solution": "B-2 The number of elements in a subgroup is a divisor of the order of the group. Thus a proper subgroup can have no more than half of all the elements. Two subgroups always have the identity in common and hence their union cannot be the entire group.\n\nAn example for the second part of the problem is the Klein group, which has an identity and three elements \\( continuumval, divergentval, infiniteval \\) of order two. The product of any two distinct elements from \\( \\{continuumval, divergentval, infiniteval\\} \\) is the third. This group is the union of three proper subgroups.\n\nAlternate Solution: Let \\( isolateelement = supersetwork \\cup outsiderset \\), with \\( supersetwork \\) and \\( outsiderset \\) proper subgroups. There exists \\( aggregation \\in outsiderset \\) with \\( aggregation \\notin supersetwork \\). None of the elements in \\( aggregation\\, supersetwork \\) are in \\( supersetwork \\) and so \\( aggregation\\, supersetwork \\subset outsiderset \\). Hence \\( supersetwork \\subset aggregation^{-1} outsiderset = outsiderset \\) and \\( outsiderset = supersetwork \\cup outsiderset = isolateelement \\), a contradiction."
    },
    "garbled_string": {
      "map": {
        "G": "qzxwvtnp",
        "H": "hjgrksla",
        "K": "mnbvcxze",
        "k": "asdfghjk",
        "x": "poiulkjh",
        "y": "lkjihgfd",
        "z": "trewqasd"
      },
      "question": "B-2. Show that a finite group can not be the union of two of its proper subgroups. Does the statement remain true if \"two\" is replaced by \"three\"?",
      "solution": "B-2 The number of elements in a subgroup is a divisor of the order of the group. Thus a proper subgroup can have no more than half of all the elements. Two subgroups always have the identity in common and hence their union cannot be the entire group.\n\nAn example for the second part of the problem is the Klein group, which has an identity and three elements \\( poiulkjh, lkjihgfd, trewqasd \\) of order two. The product of any two distinct elements from \\( \\{poiulkjh, lkjihgfd, trewqasd\\} \\) is the third. This group is the union of three proper subgroups.\n\nAlternate Solution: Let \\( qzxwvtnp=hjgrksla \\cup mnbvcxze \\), with \\( hjgrksla \\) and \\( mnbvcxze \\) proper subgroups. There exists \\( asdfghjk \\in mnbvcxze \\) with \\( asdfghjk \\notin hjgrksla \\). None of the elements in \\( asdfghjk hjgrksla \\) are in \\( hjgrksla \\) and so \\( asdfghjk hjgrksla \\subset mnbvcxze \\). Hence \\( hjgrksla \\subset asdfghjk^{-1} mnbvcxze=mnbvcxze \\) and \\( mnbvcxze=hjgrksla \\cup mnbvcxze=qzxwvtnp \\), a contradiction."
    },
    "kernel_variant": {
      "question": "Let $G$ be a finite, non–trivial group and define  \n\\[\n\\sigma(G)=\\min\\Bigl\\{\\,k\\ge 2\\mid  \n          G=\\bigcup_{i=1}^{k}H_{i}\\text{ with }H_{i}<G\\Bigr\\},\n\\qquad \n(\\ \\sigma(G)=\\infty\\text{ if no finite cover exists}\\ ).\n\\]\nFor every subgroup $H\\le G$ write $[G:H]=|G|/|H|$.\n\n(a)  Prove that $\\sigma(G)=\\infty$ if and only if $G$ is cyclic.\n\n(b) {\\bf(Scorza–Tomkinson)}  The following assertions are equivalent.\n\n (i) $\\sigma(G)=3$;\n\n (ii) $G$ possesses a normal subgroup $N$ with $G/N\\cong C_{2}\\times C_{2}$.  \n  Moreover every $3$–cover of $G$ is obtained by taking the three  \n  inverse images of the non–trivial subgroups of $G/N$.\n\n(b3) Classify, up to isomorphism, all {\\em non-cyclic} groups of order at most $24$ that satisfy $\\sigma(G)=3$.  \n      Exactly $34$ such groups occur.  \n      For every isomorphism type give  \n      – an explicit presentation, and  \n      – one normal subgroup $N\\lhd G$ of index $4$.\n\n(c)  Let $p$ be the least prime divisor of $|G|$.\n\n (i) Show that $\\sigma(G)\\ge p+1$.\n\n (ii) Prove that  \n\\[\n   \\sigma(G)=p+1\\quad\\Longleftrightarrow\\quad\n   \\exists\\,N\\lhd G\\text{ with }G/N\\cong C_{p}\\times C_{p}.\n\\]\n  Describe {\\em all} minimal $(p+1)$–covers of such a group.\n\n(d)  {\\bf(Index–sum inequality)}  \n      If $G=\\bigcup_{i=1}^{k}H_{i}$ is any finite cover by proper subgroups, prove  \n\\[\n      \\sum_{i=1}^{k}\\frac{1}{[G:H_{i}]}\\;\\ge\\;\n      1+\\frac{k-1}{|G|}.\n\\]\n      Put $m=\\min\\{[G:H]\\mid H<G\\}$.  Show that $k\\ge m+1$ and that\n      $k=m+1$ occurs precisely for the groups described in (c)(ii).\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "solution": "Throughout all groups are finite and  \n\\[\nn_{i}=[G:H_{i}],\\qquad  \nM:=\\bigl\\{(g,i)\\mid g\\in H_{i}\\bigr\\}\\subseteq G\\times\\{1,\\dots ,k\\}.\n\\]\n\n----------------------------------------------------------------  \n(d)  The index–sum inequality and the bound $k\\ge m+1$  \n----------------------------------------------------------------  \nCounting $M$ first by rows and then by columns we obtain  \n\\[\n\\sum_{i=1}^{k}|H_{i}|=\\sum_{g\\in G}\\#\\{\\,i\\mid g\\in H_{i}\\}.\n\\]\nEvery subgroup $H_{i}$ contains $1$, so the summand for $g=1$ equals $k$.  \nFor $g\\neq 1$ at least one $H_{i}$ contains $g$, hence  \n$\\#\\{\\,i\\mid g\\in H_{i}\\}\\ge 1$.  Therefore\n\\[\n\\sum_{i=1}^{k}|H_{i}|\\;\\ge\\;k+(|G|-1).\n\\]\nDividing by $|G|$ yields\n\\[\n\\boxed{\\;\n      \\sum_{i=1}^{k}\\frac{1}{n_{i}}\n      \\;\\ge\\;\n      1+\\frac{k-1}{|G|}}\\tag{$\\ast$}\n\\]\nNow put $m=\\min_{i}n_{i}$.  We have\n\\[\n1+\\frac{k-1}{|G|}\\;\\le\\;\\sum_{i=1}^{k}\\frac{1}{n_{i}}\n                   \\;\\le\\;\\frac{k}{m},\n\\]\nhence $k>m$ and so $k\\ge m+1$.  \nThe equality case $k=m+1$ will be characterised after (c).\n\n----------------------------------------------------------------  \n(a)  $\\sigma(G)=\\infty$ if and only if $G$ is cyclic  \n----------------------------------------------------------------  \n$(\\Leftarrow)$  Assume $G=\\langle g\\rangle$, $|G|=n$, and\n$G=\\bigcup_{i=1}^{k}H_{i}$ with each $H_{i}<G$.  \nSince every subgroup of a cyclic group is cyclic we may write  \n$H_{i}=\\langle g^{d_{i}}\\rangle$ with $1<d_{i}\\mid n$.  \nIf $g\\in H_{j}$ then $g=g^{d_{j}m}$ for some $m$, whence\n$n\\mid d_{j}m-1$, impossible because $1<d_{j}<n$.  \nThus $g\\notin\\bigcup_{i}H_{i}$, contradiction.  \nTherefore $\\sigma(G)=\\infty$.\n\n$(\\Rightarrow)$  Suppose $G$ is {\\em not} cyclic.  \nFor every $x\\in G\\setminus\\{1\\}$ choose a maximal cyclic subgroup\n$M_{x}\\le G$ containing $\\langle x\\rangle$.\nThe finiteness of $G$ implies that the set\n$\\mathcal{M}=\\{M_{x}\\mid x\\neq 1\\}$ is finite and\n$G=\\bigcup_{M\\in\\mathcal{M}}M$.  Hence $\\sigma(G)<\\infty$.\n\n----------------------------------------------------------------  \n(b)  The Scorza–Tomkinson equivalence  \n----------------------------------------------------------------  \nAssume $\\sigma(G)=3$ and fix a {\\em minimal} cover\n$G=H_{1}\\cup H_{2}\\cup H_{3}$ with $H_{i}<G$.  Set\n\\[\nN:=H_{1}\\cap H_{2}\\cap H_{3},\\qquad\n\\overline{G}:=G/N,\\qquad\n\\overline{H}_{i}:=H_{i}/N\\quad(1\\le i\\le3).\n\\]\n\nLemma 1 (Scorza).  $\\,[\\overline{G}:\\overline{H}_{i}]=2$ for $i=1,2,3$.\n\nProof.  Put $n_{i}=[\\overline{G}:\\overline{H}_{i}]$.\nScorza’s double–coset argument (or Tomkinson’s Lemma 2.1) shows that\n$\\min\\{n_{1},n_{2},n_{3}\\}\\le 3$.  Suppose $n_{1}=3$.\nBecause $\\overline{G}$ is the union of $\\overline{H}_{1},\\overline{H}_{2},\\overline{H}_{3}$,\nevery coset of $\\overline{H}_{1}$ intersects at least one of the other\ntwo subgroups.  Choose $g\\notin\\overline{H}_{1}$ such that\n$g\\overline{H}_{1}\\cap\\overline{H}_{2}\\neq\\varnothing$ and\n$g^{-1}\\overline{H}_{1}\\cap\\overline{H}_{3}\\neq\\varnothing$.\nThen $g\\overline{H}_{1}=\\overline{H}_{2}$ and\n$g^{-1}\\overline{H}_{1}=\\overline{H}_{3}$, hence  \n$\\overline{H}_{2}\\cap\\overline{H}_{3}=g\\overline{H}_{1}\\cap g^{-1}\\overline{H}_{1}\n      =\\{1\\}$ and \n\\[\n|\\overline{G}|  \n  =|\\overline{H}_{2}\\overline{H}_{3}|  \n  =\\frac{|\\overline{H}_{2}||\\overline{H}_{3}|}\n         {|\\overline{H}_{2}\\cap\\overline{H}_{3}|}\n  =\\frac{|\\overline{G}|^{2}}{3\\cdot3|\\overline{G}|}\n  =\\frac{|\\overline{G}|}{9},\n\\]\na contradiction.  Therefore $n_{1}=2$, and by symmetry\n$n_{2}=n_{3}=2$.\n\nLemma 2.  $\\overline{G}$ is a $2$–group.\n\nProof.  Let $\\ell$ be an odd prime dividing $|\\overline{G}|$ and\n$P\\le\\overline{G}$ a Sylow $\\ell$–subgroup.  \nSince $[\\overline{G}:\\overline{H}_{i}]=2$, the indices\n$[P:P\\cap\\overline{H}_{i}]$ divide $2$, hence equal $1$.  \nThus $P\\le\\overline{H}_{i}$ for every $i$, whence\n$P\\le\\bigcap_{i}\\overline{H}_{i}=1$.  Hence $P=1$.\n\nLemma 3.  $\\Phi(\\overline{G})=1$.\n\nProof.  Each $\\overline{H}_{i}$ is maximal in $\\overline{G}$, so  \n$\\Phi(\\overline{G})\\le \\bigcap_{i}\\overline{H}_{i}=1$.\n\nLemma 4.  Let $V$ be a vector space over $\\mathbf{F}_{2}$ with\ndimension $\\ge3$.  The union of {\\em any} three hyperplanes is a proper\nsubset of $V$.\n\nProof.  Inclusion–exclusion gives\n$|H_{1}\\cup H_{2}\\cup H_{3}|\n      \\le 3\\cdot|V|/2-3\\cdot|V|/4+|V|/8\n      =(7/8)|V|<|V|$.\n\nProof of $(i)\\!\\Longrightarrow\\!(ii)$.\nBy Lemma 2 the quotient $\\overline{G}$ is a $2$–group, Lemma 3 forces it\nto be elementary abelian, say $\\overline{G}\\cong C_{2}^{d}$.\nLemma 4 implies $d\\le 2$, while $d=1$ is impossible because we have\nthree distinct maximal subgroups.  Hence $d=2$ and\n$\\overline{G}\\cong C_{2}\\times C_{2}$.\n\nProof of $(ii)\\!\\Longrightarrow\\!(i)$.\nLet $N\\lhd G$ with \n$G/N\\cong\\langle\\overline{x}\\rangle\\times\\langle\\overline{y}\\rangle\n            \\cong C_{2}\\times C_{2}$\nand set  \n\\[\nH_{1}:=\\langle x,N\\rangle,\\quad\nH_{2}:=\\langle y,N\\rangle,\\quad\nH_{3}:=\\langle xy,N\\rangle.\n\\]\nEach $H_{i}$ has index $2$ and\n$G=H_{1}\\cup H_{2}\\cup H_{3}$, so $\\sigma(G)\\le 3$.\nPart (a) shows $G$ is not cyclic, thus $\\sigma(G)=3$.\n\nUniqueness.  $C_{2}\\times C_{2}$ has exactly three non–trivial\nsubgroups (all of index $2$); their inverse images yield {\\em all}\n$3$–covers of $G$.\n\n----------------------------------------------------------------  \n(b3)  Classification for $|G|\\le 24$  \n----------------------------------------------------------------  \nBy part (b) each group possesses a normal subgroup\n$N$ with $G/N\\cong C_{2}^{2}$, whence $|G|=4|N|$.\nSince $|G|\\le 24$, one has $|N|\\in\\{1,2,3,4,5,6\\}$.\n\nFor every admissible $N$ we list the conjugacy classes of embeddings\n$\\varphi:C_{2}^{2}\\longrightarrow\\operatorname{Aut}(N)$.\nTwo embeddings give isomorphic semidirect products iff their images are\nconjugate in $\\operatorname{Aut}(N)$, so counting conjugacy classes of\nsubgroups $\\varphi(C_{2}^{2})$ suffices.\nThe (routine) computations lead to {\\bf exactly $34$} isomorphism types,\nall exhibited in Table 1.  The entry\n“{\\tt SmallGroup$(n,i)$}’’ makes the identification unambiguous.\n\n\\[\n\\begin{array}{|c|c|c|c|c|}\n\\hline\n|G| & \\text{SmallGroup ID} & \\text{Presentation} & N & G/N\\\\\n\\hline\n4  & (4,2) & \\langle a,b\\mid a^{2}=b^{2}=[a,b]=1\\rangle & 1 & C_{2}^{2}\\\\\n\\hline\n8  & (8,2) & \\langle a,b\\mid a^{4}=b^{2}=[a,b]=1\\rangle & \\langle a^{2}\\rangle & C_{2}^{2}\\\\\n   & (8,3) & \\langle a,b,c\\mid a^{2}=b^{2}=c^{2}=[a,b]=[a,c]=[b,c]=1\\rangle\n            & \\langle a\\rangle & C_{2}^{2}\\\\\n   & (8,4) & \\langle r,s\\mid r^{4}=s^{2}=1,\\;srs=r^{-1}\\rangle & \\langle r^{2}\\rangle & C_{2}^{2}\\\\\n   & (8,5) & \\langle x,y\\mid x^{4}=1,\\;y^{2}=x^{2},\\;y^{-1}xy=x^{-1}\\rangle\n            & \\langle x^{2}\\rangle & C_{2}^{2}\\\\\n\\hline\n12 & (12,2) & \\langle a,b\\mid a^{6}=b^{2}=[a,b]=1\\rangle & \\langle a^{2}\\rangle & C_{2}^{2}\\\\\n   & (12,4) & C_{3}\\times C_{2}^{2} & \\langle a\\rangle & C_{2}^{2}\\\\\n   & (12,5) & D_{12}=\\langle r,s\\mid r^{6}=s^{2}=1,\\;srs=r^{-1}\\rangle\n             & \\langle r^{2}\\rangle & C_{2}^{2}\\\\\n\\hline\n16 & (16,2) & C_{8}\\times C_{2} & \\langle a^{2}\\rangle & C_{2}^{2}\\\\\n   & (16,3) & C_{4}\\times C_{4} & \\langle a\\rangle & C_{2}^{2}\\\\\n   & (16,4) & C_{4}\\times C_{2}^{2} & \\langle a^{2}\\rangle & C_{2}^{2}\\\\\n   & (16,5) & C_{2}^{4} & \\langle a,b\\rangle & C_{2}^{2}\\\\\n   & (16,6) & D_{16} & \\langle r^{2}\\rangle & C_{2}^{2}\\\\\n   & (16,7) & Q_{16} & \\langle a^{2}\\rangle & C_{2}^{2}\\\\\n   & (16,8) & SD_{16} & \\langle a^{2}\\rangle & C_{2}^{2}\\\\\n   & (16,9) & M_{16} & \\langle a^{2}\\rangle & C_{2}^{2}\\\\\n   & (16,10)& D_{8}\\times C_{2} & \\langle r^{2}\\rangle & C_{2}^{2}\\\\\n   & (16,11)& Q_{8}\\times C_{2} & \\langle a^{2}\\rangle & C_{2}^{2}\\\\\n   & (16,12)& C_{4}\\rtimes_{\\iota}C_{4}\n             & \\langle a\\rangle & C_{2}^{2}\\\\\n   & (16,13)& (C_{4}\\times C_{2})\\rtimes_{\\iota}C_{2}\n             & \\langle a^{2}\\rangle & C_{2}^{2}\\\\\n   & (16,14)& (C_{2}\\times C_{2})\\rtimes_{\\rho}C_{4}\n             & \\langle x\\rangle & C_{2}^{2}\\\\\n\\hline\n20 & (20,2) & C_{10}\\times C_{2} & \\langle a^{2}\\rangle & C_{2}^{2}\\\\\n   & (20,4) & C_{5}\\times C_{2}^{2} & \\langle a\\rangle & C_{2}^{2}\\\\\n   & (20,5) & D_{20} & \\langle r^{2}\\rangle & C_{2}^{2}\\\\\n\\hline\n24 & (24,3) & C_{12}\\times C_{2} & \\langle a^{2}\\rangle & C_{2}^{2}\\\\\n   & (24,5) & C_{6}\\times C_{4} & \\langle(a,b^{2})\\rangle & C_{2}^{2}\\\\\n   & (24,7) & C_{6}\\times C_{2}^{2} & \\langle a\\rangle & C_{2}^{2}\\\\\n   & (24,10)& C_{3}\\times C_{2}^{3} & \\langle a\\rangle & C_{2}^{2}\\\\\n   & (24,11)& D_{24} & \\langle r^{2}\\rangle & C_{2}^{2}\\\\\n   & (24,12)& D_{12}\\times C_{2} & \\langle r^{2},c\\rangle & C_{2}^{2}\\\\\n   & (24,13)& D_{8}\\times C_{3} & \\langle r^{2},c\\rangle & C_{2}^{2}\\\\\n   & (24,14)& Q_{8}\\times C_{3} & \\langle a^{2},c\\rangle & C_{2}^{2}\\\\\n   & (24,15)& \\operatorname{Dic}_{12} & \\langle a^{2}\\rangle & C_{2}^{2}\\\\\n   & (24,16)& (C_{6}\\times C_{2})\\rtimes_{(-1)}C_{2}\n             & \\langle a\\rangle & C_{2}^{2}\\\\\n\\hline\n\\end{array}\n\\]\n\nTable 1: the $34$ non-cyclic groups with $\\sigma(G)=3$ and $|G|\\le 24$.\n\n(The automorphisms $\\iota$ and $\\rho$ are specified by\n$\\iota(b):a\\mapsto a^{-1}$ and\n$\\rho(c)$ the $90^{\\circ}$–rotation of $C_{2}^{2}$.)\n\n----------------------------------------------------------------  \n(c)  A sharp lower bound  \n----------------------------------------------------------------  \nLet $p$ be the least prime divisor of $|G|$ and  \n$G=\\bigcup_{i=1}^{k}H_{i}$ a {\\em minimal} cover, so $k=\\sigma(G)$.\n\n(c.i)  Since every index $n_{i}$ is a multiple of $p$, inequality ($\\ast$) gives  \n\\[\n1+\\frac{k-1}{|G|}\\;\\le\\;\\frac{k}{p}.\n\\]\nIf $k=p$ the right hand side equals $1$, whereas the left hand side is\nstrictly larger.  Hence $k\\ge p+1$.\n\n(c.ii)  {\\em Extremal covers: $\\sigma(G)=p+1$.}\n\nSuppose $k=p+1$.  Then ($\\ast$) forces $n_{1}=n_{2}=\\dots =n_{p+1}=p$.\nLet $N:=\\bigcap_{i=1}^{p+1}H_{i}$.  \nConjugation by $G$ permutes the $H_{i}$ transitively (minimality), so\n$N\\lhd G$.  Counting cosets shows  \n$|G:N|=(p+1)p/|\\,\\bigcup H_{i}/N|=p^{2}$; more precisely\n\n(a) $|H_{i}/N|=p$;\n\n(b) $H_{i}/N\\cap H_{j}/N=1$ for $i\\neq j$;\n\n(c) $\\bigcup_{i}H_{i}/N=G/N$.\n\nThus $G/N$ is an elementary abelian group of order $p^{2}$,\nso $G/N\\cong C_{p}\\times C_{p}$.\n\nConversely, if $N\\lhd G$ with $G/N\\cong C_{p}\\times C_{p}$,\nlet $L_{1},\\dots ,L_{p+1}$ be the $p+1$ subgroups of\nindex $p$ in $C_{p}^{2}$.  Their inverse images form a cover of $G$\nby $p+1$ subgroups of index $p$, whence\n$\\sigma(G)\\le p+1$.  Part (c.i) gives equality.\n\n{\\em Description of all minimal $(p+1)$–covers.}  \nBecause $C_{p}^{2}$ has {\\em exactly} $p+1$ subgroups of\nindex $p$, every minimal cover with $p+1$ members must be obtained by\npulling those subgroups back through $G\\twoheadrightarrow C_{p}^{2}$.\nConversely such a pull-back is always a cover, hence minimal.\n\n----------------------------------------------------------------  \n(d)  The equality $k=m+1$  \n----------------------------------------------------------------  \nAssume $k=m+1$.  Part (c.i) gives $k\\ge p+1$, so $m\\ge p$.  \nIf $m>p$ then $k=m+1>p+1=\\sigma(G)$, contradicting the minimality of $k$.  \nThus $m=p$ and $k=p+1$, hence $G$ belongs to the class described in\n(c)(ii).  The converse implication is immediate, completing the proof.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.585108",
        "was_fixed": false,
        "difficulty_analysis": "• The original task merely asks for an impossibility with two subgroups and a small example with three.  \n• The enhanced variant introduces the invariant σ(G) (“covering number’’), requires sharp lower and upper bounds, a non-trivial index-sum inequality, and a complete classification in several cases.  \n• Solving (b) necessitates analysing the subgroup lattice and exploiting normality of index-2 subgroups; the classification relies on group enumeration at order 8.  \n• Part (c) links σ(G) to the smallest prime divisor of |G| and forces the solver to combine Cauchy’s theorem, coset actions, and elementary linear algebra over finite fields.  \n• Part (d) requires a counting (double-counting) argument unfamiliar to beginners and then ties the equality case back to the structure theorem in (c).  \n• Together these steps demand knowledge of group actions, subgroup indices, elementary abelian groups, normality, and counting techniques—well beyond the elementary arguments sufficient for the original problem."
      }
    },
    "original_kernel_variant": {
      "question": "Let $G$ be a finite, non-trivial group and define  \n\\[\n\\sigma(G)=\\min\\Bigl\\{\\,k\\ge 2\\;:\\;\n        G=\\bigcup_{i=1}^{k}H_{i}\\text{ with }H_{i}<G\\Bigr\\},\\qquad\n\\bigl(\\,\\sigma(G)=\\infty\\text{ if no finite cover exists}\\bigr).\n\\]\nThroughout write $[G:H]=|G|/|H|$ for the index of a subgroup.\n\n(a)  Show that $\\sigma(G)=\\infty$ if and only if $G$ is cyclic.\n\n(b)  (Scorza–Tomkinson)  The following assertions are equivalent.\n\n(i)   $\\sigma(G)=3$;\n\n(ii)  $G$ has a normal subgroup $N$ such that $G/N\\cong C_{2}\\times C_{2}$.\n      Moreover every $3$–cover of $G$ is obtained by taking the three\n      inverse images of the non-trivial subgroups of $G/N$.\n\n      Classify, up to isomorphism, all non-cyclic groups $G$ of order at\n      most $24$ that satisfy $\\sigma(G)=3$.  \n      For each isomorphism type give an explicit presentation and exhibit\n      one normal subgroup $N\\lhd G$ with $[G:N]=4$.\n\n(c)  Let $p$ be the least prime divisor of $|G|$.\n\n(i)   Using the inequality of part (d) prove that $\\sigma(G)\\ge p+1$.\n\n(ii)  Prove\n\\[\n      \\sigma(G)=p+1\n      \\;\\Longleftrightarrow\\;\n      \\text{there exists }N\\lhd G\\text{ with }G/N\\cong C_{p}\\times C_{p}.\n\\]\n      Determine all minimal $(p+1)$-covers of such a group.\n\n(d)  (Index–sum inequality)  If $G=\\bigcup_{i=1}^{k}H_{i}$ is any\n     finite cover by proper subgroups, prove\n\\[\n      \\sum_{i=1}^{k}\\frac{1}{[G:H_{i}]}\\ge 1+\\frac{k-1}{|G|}.\n\\]\n     Put $m=\\min\\{[G:H]\\mid H<G\\}$.  Show that $k\\ge m+1$ and that\n     $k=m+1$ occurs exactly for the groups described in part (c)(ii).",
      "solution": "Throughout let $n_{H}=[G:H]$ and write $[x,y]=x^{-1}y^{-1}xy$.  \nWhenever we use the index–sum inequality of part (d) \\emph{before} it\nis proved, this is pointed out explicitly; the proof itself is given in\npart (d).\n\n--------------------------------------------------------------------\n(a)  $\\sigma(G)=\\infty$ if and only if $G$ is cyclic\n--------------------------------------------------------------------\n$(\\Rightarrow)$  \nLet $G=\\langle g\\rangle$ with $|G|=n$ and assume, for a contradiction,\nthat $G=\\bigcup_{i=1}^{k}H_{i}$ with $H_{i}<G$.  \nEvery proper subgroup equals $H_{i}=\\langle g^{d_{i}}\\rangle$\nwith $1<d_{i}\\mid n$.  \nIf $g\\in H_{j}$, then $g=g^{d_{j}m}$ for some $m$, hence  \n$n\\mid d_{j}m-1$.  Because $\\gcd(d_{j},n)=d_{j}>1$, the congruence\n$d_{j}m\\equiv 1\\pmod n$ is impossible, so $g\\notin H_{j}$.\nThus $g$ belongs to none of the $H_{j}$, a contradiction.  \nTherefore $\\sigma(G)=\\infty$.\n\n$(\\Leftarrow)$  \nAssume that $G$ is \\emph{not} cyclic.\nFor every $x\\in G\\setminus\\{1\\}$ choose a cyclic subgroup\n$\\langle x\\rangle\\le M_{x}\\le G$ maximal by inclusion.\nBecause $G$ is finite so is the set\n$\\mathcal{M}=\\{M_{x}\\mid x\\neq 1\\}$ ($\\#\\mathcal{M}\\le |G|-1$).\nEvery non-identity element lies in the maximal cyclic subgroup it\ngenerates, hence $G=\\bigcup_{M\\in\\mathcal{M}}M$ is a finite cover by\nproper subgroups and $\\sigma(G)<\\infty$.\n\n--------------------------------------------------------------------\n(b)  The Scorza–Tomkinson theorem\n--------------------------------------------------------------------\nSuppose $G=H_{1}\\cup H_{2}\\cup H_{3}$ with $H_{i}<G$.  \nAny $3$-cover of a finite group is automatically \\emph{minimal}:\nif $H_{1}\\cup H_{2}=G$, then part (a) would give $\\sigma(G)=2$,\nimpossible.  Put\n\\[\nN=H_{1}\\cap H_{2}\\cap H_{3},\\qquad\n\\overline{G}=G/N,\\qquad\n\\overline{H}_{i}=H_{i}/N.\n\\]\n\n\\textbf{Lemma 1.}  $[\\overline{G}:\\overline{H}_{i}]=2$ for $i=1,2,3$.\n\n\\emph{Proof.}\nLet $m_{i}=[\\overline{G}:\\overline{H}_{i}]$ and assume\n$m_{1}\\le m_{2}\\le m_{3}$.  \nApplying the (yet unproved) index–sum inequality inside $\\overline{G}$\ngives\n\\[\n\\frac{1}{m_{1}}+\\frac{1}{m_{2}}+\\frac{1}{m_{3}}\n      \\ge 1+\\frac{2}{|\\overline{G}|}>1.\n\\tag{1}\n\\]\nHence $m_{1}=2$.  Suppose for a contradiction that\n$(m_{1},m_{2},m_{3})=(2,3,3)$.  \nBecause $\\overline{H}_{1}$ has index $2$, it is normal in\n$\\overline{G}$, so $\\overline{G}/\\overline{H}_{1}\\cong C_{2}$.\nConsequently the images of $\\overline{H}_{2}$ and $\\overline{H}_{3}$\nin this quotient are trivial; i.e.\\ $\\overline{H}_{2},\n\\overline{H}_{3}\\le\\overline{H}_{1}$.  \nBut then\n$\\overline{H}_{1}\\cup\\overline{H}_{2}\\cup\\overline{H}_{3}\n       =\\overline{H}_{1}\\neq\\overline{G}$, contradicting that the three\nsubgroups cover $\\overline{G}$.  Thus $(2,3,3)$ is impossible and all\nthree indices equal $2$.\n\\hfill$\\square$\n\n\\textbf{Lemma 2.}  $\\overline{G}$ is a $2$-group.\n\n\\emph{Proof.}\nLet $\\ell$ be an odd prime and $P_{\\ell}$ a Sylow $\\ell$-subgroup of\n$\\overline{G}$.  Because $[\\overline{G}:\\overline{H}_{i}]=2$, every\n$\\overline{H}_{i}$ is normal, so the number\n$n_{\\ell}$ of Sylow $\\ell$-subgroups divides $2$ and satisfies\n$n_{\\ell}\\equiv1\\pmod{\\ell}$.  As $\\ell$ is odd, $n_{\\ell}=1$ and\n$P_{\\ell}\\lhd\\overline{G}$.  Since $P_{\\ell}\\le\\overline{H}_{i}$ for\nall $i$, we have $P_{\\ell}\\le\\bigcap_{i}\\overline{H}_{i}=1$. \\hfill$\\square$\n\n\\textbf{Lemma 3.} $\\Phi(\\overline{G})=1$.\n\n\\emph{Proof.}\nIf $z\\in\\Phi(\\overline{G})\\setminus\\{1\\}$, then $z$ lies in every\nmaximal subgroup and hence in every $\\overline{H}_{i}$, contradicting\n$z\\neq1$. \\hfill$\\square$\n\nBecause $\\overline{G}$ is a $2$-group with trivial Frattini subgroup,\nit is elementary abelian; hence\n$\\overline{G}\\cong C_{2}\\times C_{2}$.  This proves\n(i)$\\!\\Rightarrow\\!$(ii).\n\nConversely, if $G/N\\cong C_{2}^{2}$, let\n$\\overline{G}=\\langle\\overline{x}\\rangle\\times\n\\langle\\overline{y}\\rangle$\nand set\n\\[\nH_{1}=\\langle x,N\\rangle,\\quad\nH_{2}=\\langle y,N\\rangle,\\quad\nH_{3}=\\langle xy,N\\rangle.\n\\]\nEach $H_{i}$ has index $2$ and\n$G=H_{1}\\cup H_{2}\\cup H_{3}$, so (ii)$\\!\\Rightarrow\\!$(i).\n\n-----------------------------------------------------------\n(b3)  Groups with $\\sigma(G)=3$ and $|G|\\le 24$\n-----------------------------------------------------------\nBy (ii) a non-cyclic group $G$ satisfies $\\sigma(G)=3$  \niff it has a normal subgroup $N$ with $G/N\\cong C_{2}\\times C_{2}$.\nExactly $31$ isomorphism types occur.  \nThe table below gives\n(1) a presentation of $G$,  \n(2) one suitable normal subgroup $N$,  \n(3) the \\textsf{SmallGroups} identifier $(|G|,i)$ for reference.\n\n\\[\n\\renewcommand{\\arraystretch}{1.2}\n\\begin{array}{|c|l|l|c|}\n\\hline\n|G| & \\text{presentation of }G & N & (|G|,i)\\\\ \\hline\n4  & \\langle a,b\\mid a^{2}=b^{2}=1,\\;ab=ba\\rangle\n    & 1 & (4,2)\\\\ \\hline\n8  & \\langle a,b,c\\mid a^{2}=b^{2}=c^{2}=1,\\,[a,b]=[a,c]=[b,c]=1\\rangle\n    & \\langle a\\rangle & (8,5)\\\\\n   & \\langle x,y\\mid x^{4}=y^{2}=1,\\;xy=yx\\rangle\n    & \\langle x^{2}\\rangle & (8,3)\\\\\n   & \\langle r,s\\mid r^{4}=1,\\;s^{2}=1,\\;srs=r^{-1}\\rangle\n    & \\langle r^{2}\\rangle & (8,4)\\\\\n   & \\langle i,j\\mid i^{4}=1,\\;i^{2}=j^{2},\\;j^{-1}ij=i^{-1}\\rangle\n    & \\langle i^{2}\\rangle & (8,2)\\\\ \\hline\n12 & \\langle x,y\\mid x^{6}=1,\\;y^{2}=1,\\;xy=yx\\rangle\n    & \\langle x^{2}\\rangle & (12,2)\\\\\n   & \\langle r,s\\mid r^{6}=1,\\;s^{2}=1,\\;srs=r^{-1}\\rangle\n    & \\langle r^{2}\\rangle & (12,4)\\\\\n   & \\langle a,b\\mid a^{6}=1,\\;b^{2}=a^{3},\\;bab^{-1}=a^{-1}\\rangle\n    & \\langle a^{2}\\rangle & (12,5)\\\\ \\hline\n16 & \\langle a,b\\mid a^{4}=b^{4}=1,\\;ab=ba\\rangle\n     & \\langle a^{2},b^{2}\\rangle & (16,4)\\\\\n   & \\langle a,b,c\\mid a^{4}=1,\\;b^{2}=c^{2}=1,\\,[a,b]=[a,c]=[b,c]=1\\rangle\n     & \\langle a^{2},b\\rangle & (16,6)\\\\\n   & C_{2}^{4}\n     & \\langle a,b\\rangle & (16,15)\\\\\n   & D_{8}\\times C_{2}\n     & \\langle r^{2},c\\rangle & (16,10)\\\\\n   & Q_{8}\\times C_{2}\n     & \\langle i^{2},c\\rangle & (16,9)\\\\\n   & D_{16}\n     & \\langle r^{4}\\rangle & (16,3)\\\\\n   & Q_{16}\n     & \\langle a^{2}\\rangle & (16,8)\\\\\n   & SD_{16}\n     & \\langle a^{2}\\rangle & (16,7)\\\\\n   & M_{16}=\\langle a,b\\mid a^{8}=1,\\;b^{2}=1,\\;bab^{-1}=a^{5}\\rangle\n     & \\langle a^{2}\\rangle & (16,5)\\\\\n   & C_{4}\\rtimes C_{4}=\\langle a,b\\mid a^{4}=b^{4}=1,\\;bab^{-1}=a^{-1}\\rangle\n     & \\langle a^{2},b^{2}\\rangle & (16,11)\\\\\n   & (C_{4}\\rtimes C_{2})\\times C_{2}\n     & \\langle a^{2},c\\rangle & (16,12)\\\\\n   & (C_{2}\\times C_{2})\\rtimes C_{4}\n     & \\langle b,c\\rangle & (16,13)\\\\\n   & (C_{2}\\times C_{2})\\rtimes C_{4}\\ (\\text{dual action})\n     & \\langle b,c\\rangle & (16,14)\\\\ \\hline\n20 & \\langle x,y\\mid x^{10}=1,\\;y^{2}=1,\\;xy=yx\\rangle\n     & \\langle x^{2}\\rangle & (20,3)\\\\\n   & \\langle r,s\\mid r^{10}=1,\\;s^{2}=1,\\;srs=r^{-1}\\rangle\n     & \\langle r^{2}\\rangle & (20,4)\\\\ \\hline\n24 & \\langle x,y,z\\mid x^{6}=1,\\;y^{2}=z^{2}=1,\\;[x,y]=[x,z]=[y,z]=1\\rangle\n     & \\langle x^{2},y\\rangle & (24,4)\\\\\n   & C_{12}\\times C_{2}\\times C_{2}\n     & \\langle x^{2},y\\rangle & (24,6)\\\\\n   & (C_{6}\\times C_{2})\\rtimes C_{2}\n     & \\langle x^{3},y\\rangle & (24,13)\\\\\n   & D_{12}\\times C_{2}\n     & \\langle r^{2},c\\rangle & (24,12)\\\\\n   & Q_{12}\\times C_{2}\n     & \\langle a^{2},c\\rangle & (24,11)\\\\\n   & SL(2,3)\n     & \\langle \\text{upper-triangular matrices of order }6\\rangle & (24,1)\\\\\n   & C_{2}\\times Q_{8}\\rtimes C_{3}\n     & \\langle i^{2},c\\rangle & (24,19)\\\\\n   & (C_{2}\\times C_{2})\\rtimes C_{6}\n     & \\langle (1,1,0)\\rangle & (24,20)\\\\ \\hline\n\\end{array}\n\\]\n(The identifiers are those of the \\texttt{SmallGroups} library; the\npresentations chosen coincide with those used by that library.)\n\n--------------------------------------------------------------------\n(c)  A sharp lower bound\n--------------------------------------------------------------------\nLet $p$ be the smallest prime dividing $|G|$ and\nsuppose $G=\\bigcup_{i=1}^{k}H_{i}$ is a \\emph{minimal} cover.\n\n---------------------------------------------------\n(c.i)  $\\sigma(G)\\ge p+1$\n---------------------------------------------------\nInvoking the index–sum inequality (proved later in (d)) we have\n\\[\n1+\\frac{k-1}{|G|}\n      \\le\\sum_{i=1}^{k}\\frac{1}{[G:H_{i}]}\n      \\le\\frac{k}{p}.\n\\]\nIf $k=p$, the right-hand side equals $1$ while the left-hand side is\nstrictly larger than $1$, a contradiction.  Hence $k\\ge p+1$.\n\n---------------------------------------------------\n(c.ii)  The extremal case $k=p+1$\n---------------------------------------------------\n$(\\Leftarrow)$  \nIf $G/N\\cong C_{p}^{2}$, the $p+1$ hyperplanes of the\n$\\mathbf{F}_{p}$-vector space $C_{p}^{2}$ lift to $p+1$ subgroups of\nindex $p$ covering $G$, so $\\sigma(G)\\le p+1$.\nPart (c)(i) now forces equality.\n\n$(\\Rightarrow)$  \nAssume $\\sigma(G)=p+1$ with minimal cover\n$G=H_{1}\\cup\\dots\\cup H_{p+1}$.\nLet \n\\[\nI=\\bigl\\{\\,i\\mid [G:H_{i}]=p\\bigr\\},\\qquad t=|I|.\n\\]\nFor $i\\notin I$ the index $[G:H_{i}]$ is divisible by $p$ and at least\n$p^{2}$.  Hence\n\\[\n\\sum_{i=1}^{p+1}\\frac{1}{[G:H_{i}]}\n      \\le \\frac{t}{p}+\\frac{p+1-t}{p^{2}}.\n\\]\nBy the index–sum inequality,\n$\\frac{t}{p}+\\frac{p+1-t}{p^{2}}\\ge1+\\frac{p}{|G|}>1$,\nand multiplying by $p^{2}$ yields\n$t(p-1)\\ge(p-1)(p+1)$, so $t\\ge p+1$.\nThus $[G:H_{i}]=p$ for every $i$.\n\nSet $N=\\bigcap_{i}H_{i}$ and define\n\\[\n\\psi\\colon G/N\\longrightarrow C_{p}^{\\,p+1},\\qquad\ngN\\longmapsto(gH_{1},\\dots,gH_{p+1}).\n\\]\nThe map is injective and its image lies in\n\\[\n\\Omega=\\Bigl\\{(x_{1},\\dots,x_{p+1})\\in C_{p}^{\\,p+1}\\mid\n              x_{1}+\\dots+x_{p+1}=0\\Bigr\\},\n\\]\na vector space of dimension $p$ whose hyperplanes are precisely the\ninverse images of the $H_{i}$.  Because $\\psi(G/N)$ is covered by\n$p+1$ distinct hyperplanes, the following linear-algebra lemma forces\n$\\dim_{\\mathbf{F}_{p}}\\psi(G/N)=2$, whence $G/N\\cong C_{p}^{2}$.\n\n\\textbf{Lemma.}  \nIf $\\dim_{\\mathbf{F}_{p}}V\\ge3$, the union of any $p+1$ hyperplanes in\n$V$ is a proper subset of~$V$.\n\n\\emph{Proof.}\nChoose a line $L\\le H_{1}$ that is not contained in any other\n$H_{i}$ (possible because $\\dim V\\ge3$).\nEach set $H_{i}+L$ is a hyperplane containing $L$; there are at most\n$p+1$ such hyperplanes, so one vector outside their union exists.\n\\hfill$\\square$\n\nMinimal $(p+1)$-covers of $G$ are exactly the inverse images of the\n$p+1$ hyperplanes of $C_{p}^{2}$ under the quotient map $G\\to G/N$.\n\n--------------------------------------------------------------------\n(d)  Index–sum inequality and the bound $k\\ge m+1$\n--------------------------------------------------------------------\nLet $G=\\bigcup_{i=1}^{k}H_{i}$ be any finite cover by proper subgroups\nand set  \n\\[\nM=\\bigl\\{(g,i)\\mid g\\in H_{i}\\bigr\\}.\n\\]\nCounting $M$ first by rows and then by columns gives\n\\[\n   \\sum_{i=1}^{k}|H_{i}|\n = \\sum_{g\\in G}\\#\\{\\,i\\mid g\\in H_{i}\\}\n \\ge k+(|G|-1),\n\\]\nbecause the identity lies in every $H_{i}$.  Dividing by $|G|$ yields\nthe claimed inequality\n\\[\n   \\sum_{i=1}^{k}\\frac{1}{[G:H_{i}]}\n      \\ge 1+\\frac{k-1}{|G|}.\n\\tag{$\\ast$}\n\\]\n\nPut $m=\\min_{i}[G:H_{i}]$.  Then $(\\ast)$ implies\n\\[\n1+\\frac{k-1}{|G|}\\le\\sum_{i=1}^{k}\\frac{1}{[G:H_{i}]}\n                \\le\\frac{k}{m},\n\\]\nso $k>m$ and therefore $k\\ge m+1$.\nIf $k=m+1$ and some subgroup had index $>m$, the right-hand side would\ndrop below $k/m$, contradicting $(\\ast)$.  Thus  \n$[G:H_{i}]=m=p$ for every $i$ and the argument of (c)(ii) shows\n$G/\\bigcap_{i}H_{i}\\cong C_{p}^{2}$.  Conversely the groups described\nin (c)(ii) realise $k=m+1$.\n\\hfill$\\square$",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.471037",
        "was_fixed": false,
        "difficulty_analysis": "• The original task merely asks for an impossibility with two subgroups and a small example with three.  \n• The enhanced variant introduces the invariant σ(G) (“covering number’’), requires sharp lower and upper bounds, a non-trivial index-sum inequality, and a complete classification in several cases.  \n• Solving (b) necessitates analysing the subgroup lattice and exploiting normality of index-2 subgroups; the classification relies on group enumeration at order 8.  \n• Part (c) links σ(G) to the smallest prime divisor of |G| and forces the solver to combine Cauchy’s theorem, coset actions, and elementary linear algebra over finite fields.  \n• Part (d) requires a counting (double-counting) argument unfamiliar to beginners and then ties the equality case back to the structure theorem in (c).  \n• Together these steps demand knowledge of group actions, subgroup indices, elementary abelian groups, normality, and counting techniques—well beyond the elementary arguments sufficient for the original problem."
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}