{
  "index": "2016-A-5",
  "type": "ALG",
  "tag": [
    "ALG",
    "COMB",
    "NT"
  ],
  "difficulty": "",
  "question": "Suppose that $G$ is a finite group generated by the two elements $g$ and $h$, where the order of $g$ is odd. Show that every element of $G$ can be written in the form\n\\[\ng^{m_1} h^{n_1} g^{m_2} h^{n_2} \\cdots g^{m_r} h^{n_r}\n\\]\nwith $1 \\leq r \\leq |G|$ and $m_1, n_1, m_2, n_2, \\ldots, m_r, n_r \\in \\{-1, 1\\}$. \n(Here $|G|$ is the number of elements of $G$.)\n\n\\,",
  "solution": "\\noindent\n\\textbf{First solution:}\nFor $s \\in G$ and $r$ a positive integer, define a \\emph{representation of $s$ of length $r$}\nto be a sequence of values $m_1, n_1, \\ldots, m_r, n_r \\in \\{-1, 1\\}$ for which\n\\[\ns = g^{m_1} h^{n_1} \\cdots g^{m_r} h^{n_r}.\n\\]\nWe first check that every $s \\in G$ admits at least one representation of some length; this is equivalent to saying that the set $S$ of $s \\in G$ which admit representations of some length\nis equal to $G$ itself. \nSince $S$ is closed under the group operation and $G$ is finite, $S$ is also closed under formation of inverses and contains the identity element; that is, $S$ is a subgroup of $G$.\nIn particular, $S$ contains not only $gh$ but also its inverse $h^{-1} g^{-1}$; since $S$ also contains $g^{-1} h$, we deduce that $S$ contains $g^{-2}$. \nSince $g$ is of odd order in $G$, $g^{-2}$ is also a generator of the cyclic subgroup containing $g$; it follows that $g \\in S$ and hence $h \\in S$. Since we assumed that $g,h$ generate $G$,\nwe now conclude that $S = G$, as claimed.\n\nTo complete the proof, we must now check that for each $s \\in G$, the smallest possible length of a representation of $s$ cannot exceed $|G|$. Suppose the contrary, and let\n\\[\ns = g^{m_1} h^{n_1} \\cdots g^{m_r} h^{n_r}\n\\]\nbe a representation of the smallest possible length. Set\n\\[\ns_i = g^{m_1} h^{n_1} \\cdots g^{m_i} h^{n_i} \\qquad (i=0,\\dots,r-1),\n\\]\ninterpreting $s_0$ as $e$; since $r>|G|$ by hypothesis, by the pigeonhole principle there must exist indices $0 \\leq i < j \\leq r-1$ such that $s_i = s_j$. Then\n\\[\ns = g^{m_1} h^{n_1} \\cdots g^{m_i} h^{n_i} g^{m_{j+1}} h^{n_{j+1}} \\cdots g^{m_r} h^{n_r} \n\\]\nis another representation of $s$ of length strictly less than $r$, a contradiction.\n\n\\noindent\n\\textbf{Remark:}\nIf one considers $s_1,\\dots,s_r$ instead of $s_0,\\dots,s_{r-1}$, then the case $s=e$ must be handled separately: otherwise, one might end up with a representation of length 0 which is disallowed by the problem statement.\n\n\\noindent\n\\textbf{Reinterpretation:} \nNote that the elements $gh, gh^{-1}, g^{-1} h, g^{-1} h^{-1}$ generate $gh(g^{-1}h)^{-1} = g^2$ and hence all of $G$ (again using the hypothesis that $g$ has odd order, as above). Form the Cayley digraph on the set $G$, i.e., the directed graph with an edge from $s_1$ to $s_2$ whenever $s_2 = s_1 *$ for $* \\in \\{gh, gh^{-1}, g^{-1} h, g^{-1} h^{-1}\\}$. Since $G$ is finite, this digraph is strongly connected: there exists at least one path from any vertex to any other vertex (traveling all edges in the correct direction). The shortest such path cannot repeat any vertices (except the starting and ending vertices in case they coincide), and so has length at most $|G|$.\n\n\\noindent\n\\textbf{Second solution:}\nFor $r$ a positive integer, let $S_r$ be the set of $s \\in G$ which admit a representation of length at most $r$ (terminology as in the first solution); obviously $S_r \\subseteq S_{r+1}$.\nWe will show that $S_r \\neq S_{r+1}$ unless $S_r = G$; this will imply by induction on $r$ that $\\#S_r \\geq \\min\\{r, |G|\\}$ and hence that $S_r = G$ for some $r \\leq |G|$.\n\nSuppose that $S_r = S_{r+1}$.\nThen the map $s \\mapsto sgh$ defines an injective map $S_r \\to S_{r+1} = S_r$,\nso $S_r$ is closed under right multiplication by $gh$. By the same token, $S_r$ is closed under right multiplication by each of $gh^{-1}, g^{-1}h, g^{-1}h^{-1}$.\nSince $gh, gh^{-1}, g^{-1}h, g^{-1}h^{-1}$ generate $G$ as in the first solution, it follows that $S_r = G$ as claimed.\n\n\\noindent\n\\textbf{Remark:} The condition on the order of $g$ is needed to rule out the case where $G$ admits a (necessarily normal) subgroup $H$ of index 2 not containing either $g$ or $h$; in this case, all products of the indicated form belong to $H$.\nOn the other hand, if one assumes that both $g$ and $h$ have odd order, then one can say a bit more: there exists some positive integer $r$ with $1 \\leq r \\leq |G|$ such that every element of $G$ has a representation of length exactly $r$. (Namely, the set of such elements for a given $r$ strictly increases in size until it is stable under right multiplication by both\n$gh(g^{-1}h)^{-1} = g^2$ and $gh(gh^{-1})^{-1} = gh^2g^{-1}$, but under the present hypotheses these generate $G$.)",
  "vars": [
    "s",
    "r",
    "m",
    "n",
    "i",
    "j",
    "S",
    "m_1",
    "n_1",
    "m_2",
    "n_2",
    "m_r",
    "n_r",
    "s_i",
    "s_j",
    "s_0",
    "s_1",
    "s_r",
    "S_r"
  ],
  "params": [
    "G",
    "g",
    "h",
    "e",
    "H"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "s": "elementvar",
        "r": "lengthvar",
        "m": "exponentm",
        "n": "exponentn",
        "i": "indexone",
        "j": "indextwo",
        "S": "subsetvar",
        "m_1": "firstmex",
        "n_1": "firstnex",
        "m_2": "secondmex",
        "n_2": "secondnex",
        "m_r": "lastmex",
        "n_r": "lastnex",
        "s_i": "seqelemone",
        "s_j": "seqelemtwo",
        "s_0": "seqzero",
        "s_1": "seqone",
        "s_r": "seqlast",
        "S_r": "subsetlen",
        "G": "groupall",
        "g": "genfirst",
        "h": "gensecond",
        "e": "identity",
        "H": "subgroup"
      },
      "question": "Suppose that $groupall$ is a finite group generated by the two elements $genfirst$ and $gensecond$, where the order of $genfirst$ is odd. Show that every element of $groupall$ can be written in the form\n\\[\ngenfirst^{firstmex} gensecond^{firstnex} genfirst^{secondmex} gensecond^{secondnex} \\cdots genfirst^{lastmex} gensecond^{lastnex}\n\\]\nwith $1 \\leq lengthvar \\leq |groupall|$ and $firstmex, firstnex, secondmex, secondnex, \\ldots, lastmex, lastnex \\in \\{-1, 1\\}$. \n(Here $|groupall|$ is the number of elements of $groupall$.)",
      "solution": "\\noindent\n\\textbf{First solution:}\nFor $elementvar \\in groupall$ and $lengthvar$ a positive integer, define a \\emph{representation of $elementvar$ of length $lengthvar$}\nto be a sequence of values $firstmex, firstnex, \\ldots, lastmex, lastnex \\in \\{-1, 1\\}$ for which\n\\[\nelementvar = genfirst^{firstmex} gensecond^{firstnex} \\cdots genfirst^{lastmex} gensecond^{lastnex}.\n\\]\nWe first check that every $elementvar \\in groupall$ admits at least one representation of some length; this is equivalent to saying that the set $subsetvar$ of $elementvar \\in groupall$ which admit representations of some length\nis equal to $groupall$ itself. \nSince $subsetvar$ is closed under the group operation and $groupall$ is finite, $subsetvar$ is also closed under formation of inverses and contains the identity element; that is, $subsetvar$ is a subgroup of $groupall$.\nIn particular, $subsetvar$ contains not only $genfirstgensecond$ but also its inverse $gensecond^{-1} genfirst^{-1}$; since $subsetvar$ also contains $genfirst^{-1} gensecond$, we deduce that $subsetvar$ contains $genfirst^{-2}$. \nSince $genfirst$ is of odd order in $groupall$, $genfirst^{-2}$ is also a generator of the cyclic subgroup containing $genfirst$; it follows that $genfirst \\in subsetvar$ and hence $gensecond \\in subsetvar$. Since we assumed that $genfirst,gensecond$ generate $groupall$,\nwe now conclude that $subsetvar = groupall$, as claimed.\n\nTo complete the proof, we must now check that for each $elementvar \\in groupall$, the smallest possible length of a representation of $elementvar$ cannot exceed $|groupall|$. Suppose the contrary, and let\n\\[\nelementvar = genfirst^{firstmex} gensecond^{firstnex} \\cdots genfirst^{lastmex} gensecond^{lastnex}\n\\]\nbe a representation of the smallest possible length. Set\n\\[\nseqelemone = genfirst^{firstmex} gensecond^{firstnex} \\cdots genfirst^{exponentm_{indexone}} gensecond^{exponentn_{indexone}} \\qquad (indexone=0,\\dots,lengthvar-1),\n\\]\ninterpreting $seqzero$ as $identity$; since $lengthvar>|groupall|$ by hypothesis, by the pigeonhole principle there must exist indices $0 \\leq indexone < indextwo \\leq lengthvar-1$ such that $seqelemone = seqelemtwo$. Then\n\\[\nelementvar = genfirst^{firstmex} gensecond^{firstnex} \\cdots genfirst^{exponentm_{indexone}} gensecond^{exponentn_{indexone}} genfirst^{exponentm_{indextwo+1}} gensecond^{exponentn_{indextwo+1}} \\cdots genfirst^{lastmex} gensecond^{lastnex} \n\\]\nis another representation of $elementvar$ of length strictly less than $lengthvar$, a contradiction.\n\n\\noindent\n\\textbf{Remark:}\nIf one considers $seqone,\\dots,seqlast$ instead of $seqzero,\\dots,elementvar_{lengthvar-1}$, then the case $elementvar=identity$ must be handled separately: otherwise, one might end up with a representation of length 0 which is disallowed by the problem statement.\n\n\\noindent\n\\textbf{Reinterpretation:} \nNote that the elements $genfirstgensecond, genfirstgensecond^{-1}, genfirst^{-1} gensecond, genfirst^{-1} gensecond^{-1}$ generate $genfirstgensecond(genfirst^{-1}gensecond)^{-1} = genfirst^2$ and hence all of $groupall$ (again using the hypothesis that $genfirst$ has odd order, as above). Form the Cayley digraph on the set $groupall$, i.e., the directed graph with an edge from $seqone$ to $elementvar_2$ whenever $elementvar_2 = seqone *$ for $* \\in \\{genfirstgensecond, genfirstgensecond^{-1}, genfirst^{-1} gensecond, genfirst^{-1} gensecond^{-1}\\}$. Since $groupall$ is finite, this digraph is strongly connected: there exists at least one path from any vertex to any other vertex (traveling all edges in the correct direction). The shortest such path cannot repeat any vertices (except the starting and ending vertices in case they coincide), and so has length at most $|groupall|$.\n\n\\noindent\n\\textbf{Second solution:}\nFor $lengthvar$ a positive integer, let $subsetlen$ be the set of $elementvar \\in groupall$ which admit a representation of length at most $lengthvar$ (terminology as in the first solution); obviously $subsetlen \\subseteq subsetvar_{lengthvar+1}$.\nWe will show that $subsetlen \\neq subsetvar_{lengthvar+1}$ unless $subsetlen = groupall$; this will imply by induction on $lengthvar$ that $\\#subsetlen \\geq \\min\\{lengthvar, |groupall|\\}$ and hence that $subsetlen = groupall$ for some $lengthvar \\leq |groupall|$.\n\nSuppose that $subsetlen = subsetvar_{lengthvar+1}$.\nThen the map $elementvar \\mapsto elementvar genfirstgensecond$ defines an injective map $subsetlen \\to subsetvar_{lengthvar+1} = subsetlen$,\nso $subsetlen$ is closed under right multiplication by $genfirstgensecond$. By the same token, $subsetlen$ is closed under right multiplication by each of $genfirstgensecond^{-1}, genfirst^{-1}gensecond, genfirst^{-1}gensecond^{-1}$.\nSince $genfirstgensecond, genfirstgensecond^{-1}, genfirst^{-1}gensecond, genfirst^{-1}gensecond^{-1}$ generate $groupall$ as in the first solution, it follows that $subsetlen = groupall$ as claimed.\n\n\\noindent\n\\textbf{Remark:} The condition on the order of $genfirst$ is needed to rule out the case where $groupall$ admits a (necessarily normal) subgroup $subgroup$ of index 2 not containing either $genfirst$ or $gensecond$; in this case, all products of the indicated form belong to $subgroup$.\nOn the other hand, if one assumes that both $genfirst$ and $gensecond$ have odd order, then one can say a bit more: there exists some positive integer $lengthvar$ with $1 \\leq lengthvar \\leq |groupall|$ such that every element of $groupall$ has a representation of length exactly $lengthvar$. (Namely, the set of such elements for a given $lengthvar$ strictly increases in size until it is stable under right multiplication by both\n$genfirstgensecond(genfirst^{-1}gensecond)^{-1} = genfirst^2$ and $genfirstgensecond(genfirstgensecond^{-1})^{-1} = genfirstgensecond^2genfirst^{-1}$, but under the present hypotheses these generate $groupall$.)"
    },
    "descriptive_long_confusing": {
      "map": {
        "s": "elephant",
        "r": "pineapple",
        "m": "notebook",
        "n": "butterfly",
        "i": "kangaroo",
        "j": "strawberry",
        "S": "sandwich",
        "m_1": "telescope",
        "n_1": "chocolate",
        "m_2": "waterfall",
        "n_2": "marshmallow",
        "m_r": "horsepower",
        "n_r": "skateboard",
        "s_i": "carousel",
        "s_j": "lighthouse",
        "s_0": "cinnamon",
        "s_1": "porcupine",
        "s_r": "armadillo",
        "S_r": "jellyfish",
        "G": "adventure",
        "g": "spaghetti",
        "h": "umbrella",
        "e": "sapphire",
        "H": "orchestra"
      },
      "question": "Suppose that $adventure$ is a finite group generated by the two elements $spaghetti$ and $umbrella$, where the order of $spaghetti$ is odd. Show that every element of $adventure$ can be written in the form\n\\[\nspaghetti^{telescope} umbrella^{chocolate} spaghetti^{waterfall} umbrella^{marshmallow} \\cdots spaghetti^{horsepower} umbrella^{skateboard}\n\\]\nwith $1 \\leq pineapple \\leq |adventure|$ and $telescope, chocolate, waterfall, marshmallow, \\ldots, horsepower, skateboard \\in \\{-1, 1\\}$. \n(Here $|adventure|$ is the number of elements of $adventure$.)\n",
      "solution": "\\noindent\n\\textbf{First solution:}\nFor $elephant \\in adventure$ and $pineapple$ a positive integer, define a \\emph{representation of $elephant$ of length $pineapple$}\nto be a sequence of values $telescope, chocolate, \\ldots, horsepower, skateboard \\in \\{-1, 1\\}$ for which\n\\[\nelephant = spaghetti^{telescope} umbrella^{chocolate} \\cdots spaghetti^{horsepower} umbrella^{skateboard}.\n\\]\nWe first check that every $elephant \\in adventure$ admits at least one representation of some length; this is equivalent to saying that the set $sandwich$ of $elephant \\in adventure$ which admit representations of some length\nis equal to $adventure$ itself. \nSince $sandwich$ is closed under the group operation and adventure is finite, $sandwich$ is also closed under formation of inverses and contains the identity element; that is, $sandwich$ is a subgroup of adventure.\nIn particular, $sandwich$ contains not only $spaghetti umbrella$ but also its inverse $umbrella^{-1} spaghetti^{-1}$; since $sandwich$ also contains $spaghetti^{-1} umbrella$, we deduce that $sandwich$ contains $spaghetti^{-2}$. \nSince $spaghetti$ is of odd order in adventure, $spaghetti^{-2}$ is also a generator of the cyclic subgroup containing $spaghetti$; it follows that $spaghetti \\in sandwich$ and hence $umbrella \\in sandwich$. Since we assumed that $spaghetti,umbrella$ generate adventure,\nwe now conclude that $sandwich = adventure$, as claimed.\n\nTo complete the proof, we must now check that for each $elephant \\in adventure$, the smallest possible length of a representation of $elephant$ cannot exceed $|adventure|$. Suppose the contrary, and let\n\\[\nelephant = spaghetti^{telescope} umbrella^{chocolate} \\cdots spaghetti^{horsepower} umbrella^{skateboard}\n\\]\nbe a representation of the smallest possible length. Set\n\\[\ncarousel = spaghetti^{telescope} umbrella^{chocolate} \\cdots spaghetti^{notebook_{kangaroo}} umbrella^{butterfly_{kangaroo}} \\qquad (kangaroo=0,\\dots,pineapple-1),\n\\]\ninterpreting $cinnamon$ as $sapphire$; since $pineapple>|adventure|$ by hypothesis, by the pigeonhole principle there must exist indices $0 \\leq kangaroo < strawberry \\leq pineapple-1$ such that $carousel = lighthouse$. Then\n\\[\nelephant = spaghetti^{telescope} umbrella^{chocolate} \\cdots spaghetti^{notebook_{kangaroo}} umbrella^{butterfly_{kangaroo}} spaghetti^{notebook_{strawberry+1}} umbrella^{butterfly_{strawberry+1}} \\cdots spaghetti^{horsepower} umbrella^{skateboard} \n\\]\nis another representation of $elephant$ of length strictly less than $pineapple$, a contradiction.\n\n\\noindent\n\\textbf{Remark:}\nIf one considers $porcupine,\\dots,armadillo$ instead of $cinnamon,\\dots,carousel$, then the case $elephant=sapphire$ must be handled separately: otherwise, one might end up with a representation of length 0 which is disallowed by the problem statement.\n\n\\noindent\n\\textbf{Reinterpretation:} \nNote that the elements $spaghetti umbrella, spaghetti umbrella^{-1}, spaghetti^{-1} umbrella, spaghetti^{-1} umbrella^{-1}$ generate $spaghetti umbrella(spaghetti^{-1} umbrella)^{-1} = spaghetti^{2}$ and hence all of adventure (again using the hypothesis that $spaghetti$ has odd order, as above). Form the Cayley digraph on the set adventure, i.e., the directed graph with an edge from $elephant_1$ to $elephant_2$ whenever $elephant_2 = elephant_1 *$ for $* \\in \\{spaghetti umbrella, spaghetti umbrella^{-1}, spaghetti^{-1} umbrella, spaghetti^{-1} umbrella^{-1}\\}$. Since adventure is finite, this digraph is strongly connected: there exists at least one path from any vertex to any other vertex (traveling all edges in the correct direction). The shortest such path cannot repeat any vertices (except the starting and ending vertices in case they coincide), and so has length at most $|adventure|$.\n\n\\noindent\n\\textbf{Second solution:}\nFor $pineapple$ a positive integer, let $jellyfish$ be the set of $elephant \\in adventure$ which admit a representation of length at most $pineapple$ (terminology as in the first solution); obviously $jellyfish \\subseteq sandwich_{pineapple+1}$.\nWe will show that $jellyfish \\neq sandwich_{pineapple+1}$ unless $jellyfish = adventure$; this will imply by induction on $pineapple$ that $\\#jellyfish \\geq \\min\\{pineapple, |adventure|\\}$ and hence that $jellyfish = adventure$ for some $pineapple \\leq |adventure|$.\n\nSuppose that $jellyfish = sandwich_{pineapple+1}$.\nThen the map $elephant \\mapsto elephant spaghetti umbrella$ defines an injective map $jellyfish \\to sandwich_{pineapple+1} = jellyfish$,\nso $jellyfish$ is closed under right multiplication by $spaghetti umbrella$. By the same token, $jellyfish$ is closed under right multiplication by each of $spaghetti umbrella^{-1}, spaghetti^{-1} umbrella, spaghetti^{-1} umbrella^{-1}$.\nSince $spaghetti umbrella, spaghetti umbrella^{-1}, spaghetti^{-1} umbrella, spaghetti^{-1} umbrella^{-1}$ generate adventure as in the first solution, it follows that $jellyfish = adventure$ as claimed.\n\n\\noindent\n\\textbf{Remark:} The condition on the order of $spaghetti$ is needed to rule out the case where adventure admits a (necessarily normal) subgroup orchestra of index 2 not containing either $spaghetti$ or $umbrella$; in this case, all products of the indicated form belong to orchestra.\nOn the other hand, if one assumes that both $spaghetti$ and $umbrella$ have odd order, then one can say a bit more: there exists some positive integer $pineapple$ with $1 \\leq pineapple \\leq |adventure|$ such that every element of adventure has a representation of length exactly $pineapple$. (Namely, the set of such elements for a given $pineapple$ strictly increases in size until it is stable under right multiplication by both\n$spaghetti umbrella(spaghetti^{-1} umbrella)^{-1} = spaghetti^{2}$ and $spaghetti umbrella(spaghetti umbrella^{-1})^{-1} = spaghetti umbrella^{2} spaghetti^{-1}$, but under the present hypotheses these generate adventure.)"
    },
    "descriptive_long_misleading": {
      "map": {
        "s": "nonmember",
        "r": "fullbreadth",
        "m": "dividerer",
        "n": "unifierr",
        "j": "smallindex",
        "S": "emptyset",
        "m_1": "dividerone",
        "n_1": "unifierone",
        "m_2": "dividertwo",
        "n_2": "unifiertwo",
        "m_r": "dividerlast",
        "n_r": "unifierlast",
        "s_i": "nonmemberi",
        "s_j": "nonmemberj",
        "s_0": "nonmemberzero",
        "s_1": "nonmemberone",
        "s_r": "nonmemberr",
        "S_r": "emptysetr",
        "G": "randomset",
        "g": "destroyer",
        "h": "obliterator",
        "H": "wholegroup"
      },
      "question": "Suppose that $randomset$ is a finite group generated by the two elements $destroyer$ and $obliterator$, where the order of $destroyer$ is odd. Show that every element of $randomset$ can be written in the form\n\\[\ndestroyer^{dividerone} obliterator^{unifierone} destroyer^{dividertwo} obliterator^{unifiertwo} \\cdots destroyer^{dividerlast} obliterator^{unifierlast}\n\\]\nwith $1 \\leq fullbreadth \\leq |randomset|$ and $dividerone,\\, unifierone,\\, dividertwo,\\, unifiertwo,\\, \\ldots,\\, dividerlast,\\, unifierlast \\in \\{-1, 1\\}$. \n(Here $|randomset|$ is the number of elements of $randomset$.)",
      "solution": "\\noindent\n\\textbf{First solution:}\nFor $nonmember \\in randomset$ and $fullbreadth$ a positive integer, define a \\emph{representation of $nonmember$ of length $fullbreadth$}\nto be a sequence of values $dividerone, unifierone, \\ldots, dividerlast, unifierlast \\in \\{-1, 1\\}$ for which\n\\[\nnonmember = destroyer^{dividerone} obliterator^{unifierone} \\cdots destroyer^{dividerlast} obliterator^{unifierlast}.\n\\]\nWe first check that every $nonmember \\in randomset$ admits at least one representation of some length; this is equivalent to saying that the set $emptyset$ of $nonmember \\in randomset$ which admit representations of some length\nis equal to $randomset$ itself. \nSince $emptyset$ is closed under the group operation and $randomset$ is finite, $emptyset$ is also closed under formation of inverses and contains the identity element; that is, $emptyset$ is a subgroup of $randomset$.\nIn particular, $emptyset$ contains not only $destroyerobliterator$ but also its inverse $obliterator^{-1} destroyer^{-1}$; since $emptyset$ also contains $destroyer^{-1} obliterator$, we deduce that $emptyset$ contains $destroyer^{-2}$. \nSince $destroyer$ is of odd order in $randomset$, $destroyer^{-2}$ is also a generator of the cyclic subgroup containing $destroyer$; it follows that $destroyer \\in emptyset$ and hence $obliterator \\in emptyset$. Since we assumed that $destroyer,obliterator$ generate $randomset$,\nwe now conclude that $emptyset = randomset$, as claimed.\n\nTo complete the proof, we must now check that for each $nonmember \\in randomset$, the smallest possible length of a representation of $nonmember$ cannot exceed $|randomset|$. Suppose the contrary, and let\n\\[\nnonmember = destroyer^{dividerone} obliterator^{unifierone} \\cdots destroyer^{dividerlast} obliterator^{unifierlast}\n\\]\nbe a representation of the smallest possible length. Set\n\\[\nnonmemberi = destroyer^{dividerone} obliterator^{unifierone} \\cdots destroyer^{m_i} obliterator^{n_i} \\qquad (i=0,\\dots,fullbreadth-1),\n\\]\ninterpreting $nonmemberzero$ as $e$; since $fullbreadth>|randomset|$ by hypothesis, by the pigeonhole principle there must exist indices $0 \\leq i < smallindex \\leq fullbreadth-1$ such that $nonmemberi = nonmemberj$. Then\n\\[\nnonmember = destroyer^{dividerone} obliterator^{unifierone} \\cdots destroyer^{m_i} obliterator^{n_i} destroyer^{m_{smallindex+1}} obliterator^{n_{smallindex+1}} \\cdots destroyer^{dividerlast} obliterator^{unifierlast} \n\\]\nis another representation of $nonmember$ of length strictly less than $fullbreadth$, a contradiction.\n\n\\noindent\n\\textbf{Remark:}\nIf one considers $nonmemberone,\\dots,nonmemberr$ instead of $nonmemberzero,\\dots,nonmember_{fullbreadth-1}$, then the case $nonmember=e$ must be handled separately: otherwise, one might end up with a representation of length 0 which is disallowed by the problem statement.\n\n\\noindent\n\\textbf{Reinterpretation:} \nNote that the elements $destroyerobliterator, destroyerobliterator^{-1}, destroyer^{-1} obliterator, destroyer^{-1} obliterator^{-1}$ generate $destroyerobliterator(destroyer^{-1}obliterator)^{-1} = destroyer^{2}$ and hence all of $randomset$ (again using the hypothesis that $destroyer$ has odd order, as above). Form the Cayley digraph on the set $randomset$, i.e., the directed graph with an edge from $nonmemberone$ to $nonmember_{2}$ whenever $nonmember_{2} = nonmemberone *$ for $* \\in \\{destroyerobliterator, destroyerobliterator^{-1}, destroyer^{-1} obliterator, destroyer^{-1} obliterator^{-1}\\}$. Since $randomset$ is finite, this digraph is strongly connected: there exists at least one path from any vertex to any other vertex (traveling all edges in the correct direction). The shortest such path cannot repeat any vertices (except the starting and ending vertices in case they coincide), and so has length at most $|randomset|$.\n\n\\noindent\n\\textbf{Second solution:}\nFor $fullbreadth$ a positive integer, let $emptysetr$ be the set of $nonmember \\in randomset$ which admit a representation of length at most $fullbreadth$ (terminology as in the first solution); obviously $emptysetr \\subseteq emptyset_{fullbreadth+1}$.\nWe will show that $emptysetr \\neq emptyset_{fullbreadth+1}$ unless $emptysetr = randomset$; this will imply by induction on $fullbreadth$ that $\\#emptysetr \\geq \\min\\{fullbreadth, |randomset|\\}$ and hence that $emptysetr = randomset$ for some $fullbreadth \\leq |randomset|$.\n\nSuppose that $emptysetr = emptyset_{fullbreadth+1}$.\nThen the map $nonmember \\mapsto nonmember destroyerobliterator$ defines an injective map $emptysetr \\to emptyset_{fullbreadth+1} = emptysetr$,\nso $emptysetr$ is closed under right multiplication by $destroyerobliterator$. By the same token, $emptysetr$ is closed under right multiplication by each of $destroyerobliterator^{-1}, destroyer^{-1}obliterator, destroyer^{-1}obliterator^{-1}$.\nSince $destroyerobliterator, destroyerobliterator^{-1}, destroyer^{-1}obliterator, destroyer^{-1}obliterator^{-1}$ generate $randomset$ as in the first solution, it follows that $emptysetr = randomset$ as claimed.\n\n\\noindent\n\\textbf{Remark:} The condition on the order of $destroyer$ is needed to rule out the case where $randomset$ admits a (necessarily normal) subgroup $wholegroup$ of index 2 not containing either $destroyer$ or $obliterator$; in this case, all products of the indicated form belong to $wholegroup$.\nOn the other hand, if one assumes that both $destroyer$ and $obliterator$ have odd order, then one can say a bit more: there exists some positive integer $fullbreadth$ with $1 \\leq fullbreadth \\leq |randomset|$ such that every element of $randomset$ has a representation of length exactly $fullbreadth$. (Namely, the set of such elements for a given $fullbreadth$ strictly increases in size until it is stable under right multiplication by both\n$destroyerobliterator(destroyer^{-1}obliterator)^{-1} = destroyer^{2}$ and $destroyerobliterator(destroyer\\, obliterator^{-1})^{-1} = destroyer\\, obliterator^{2} destroyer^{-1}$, but under the present hypotheses these generate $randomset$.)"
    },
    "garbled_string": {
      "map": {
        "s": "qzxwvtnp",
        "r": "lskdjfgh",
        "m": "asqwerty",
        "n": "zmbxcvnl",
        "i": "poiuylkj",
        "j": "trewqasd",
        "S": "vbnmasdf",
        "m_1": "ghjklpoi",
        "n_1": "qazxswec",
        "m_2": "rfvtgbnh",
        "n_2": "yhnujmki",
        "m_r": "uiojklmn",
        "n_r": "plmkoijn",
        "s_i": "bfhgtrde",
        "s_j": "cdrfvtgb",
        "s_0": "edcbgfra",
        "s_1": "tgbnhyuj",
        "s_r": "wsxedcrf",
        "S_r": "nhygtvfr",
        "G": "qwertyop",
        "g": "asdfghjk",
        "h": "zxcvbnml",
        "e": "poiuytre",
        "H": "lkjhgfds"
      },
      "question": "Suppose that $qwertyop$ is a finite group generated by the two elements $asdfghjk$ and $zxcvbnml$, where the order of $asdfghjk$ is odd. Show that every element of $qwertyop$ can be written in the form\n\\[\nasdfghjk^{ghjklpoi} zxcvbnml^{qazxswec} asdfghjk^{rfvtgbnh} zxcvbnml^{yhnujmki} \\cdots asdfghjk^{uiojklmn} zxcvbnml^{plmkoijn}\n\\]\nwith $1 \\leq lskdjfgh \\leq |qwertyop|$ and $ghjklpoi, qazxswec, rfvtgbnh, yhnujmki, \\ldots, uiojklmn, plmkoijn \\in \\{-1, 1\\}$. \n(Here $|qwertyop|$ is the number of elements of $qwertyop$.)",
      "solution": "\\noindent\n\\textbf{First solution:}\nFor $qzxwvtnp \\in qwertyop$ and $lskdjfgh$ a positive integer, define a \\emph{representation of $qzxwvtnp$ of length $lskdjfgh$}\nto be a sequence of values $ghjklpoi, qazxswec, \\ldots, uiojklmn, plmkoijn \\in \\{-1, 1\\}$ for which\n\\[\nqzxwvtnp = asdfghjk^{ghjklpoi} zxcvbnml^{qazxswec} \\cdots asdfghjk^{uiojklmn} zxcvbnml^{plmkoijn}.\n\\]\nWe first check that every $qzxwvtnp \\in qwertyop$ admits at least one representation of some length; this is equivalent to saying that the set $vbnmasdf$ of $qzxwvtnp \\in qwertyop$ which admit representations of some length\nis equal to $qwertyop$ itself. \nSince $vbnmasdf$ is closed under the group operation and $qwertyop$ is finite, $vbnmasdf$ is also closed under formation of inverses and contains the identity element; that is, $vbnmasdf$ is a subgroup of $qwertyop$.\nIn particular, $vbnmasdf$ contains not only $asdfghjk zxcvbnml$ but also its inverse $zxcvbnml^{-1} asdfghjk^{-1}$; since $vbnmasdf$ also contains $asdfghjk^{-1} zxcvbnml$, we deduce that $vbnmasdf$ contains $asdfghjk^{-2}$. \nSince $asdfghjk$ is of odd order in $qwertyop$, $asdfghjk^{-2}$ is also a generator of the cyclic subgroup containing $asdfghjk$; it follows that $asdfghjk \\in vbnmasdf$ and hence $zxcvbnml \\in vbnmasdf$. Since we assumed that $asdfghjk,zxcvbnml$ generate $qwertyop$,\nwe now conclude that $vbnmasdf = qwertyop$, as claimed.\n\nTo complete the proof, we must now check that for each $qzxwvtnp \\in qwertyop$, the smallest possible length of a representation of $qzxwvtnp$ cannot exceed $|qwertyop|$. Suppose the contrary, and let\n\\[\nqzxwvtnp = asdfghjk^{ghjklpoi} zxcvbnml^{qazxswec} \\cdots asdfghjk^{uiojklmn} zxcvbnml^{plmkoijn}\n\\]\nbe a representation of the smallest possible length. Set\n\\[\nbfhgtrde = asdfghjk^{ghjklpoi} zxcvbnml^{qazxswec} \\cdots asdfghjk^{asqwerty_{poiuylkj}} zxcvbnml^{zmbxcvnl_{poiuylkj}} \\qquad (poiuylkj=0,\\dots,lskdjfgh-1),\n\\]\ninterpreting $edcbgfra$ as $poiuytre$; since $lskdjfgh>|qwertyop|$ by hypothesis, by the pigeonhole principle there must exist indices $0 \\leq poiuylkj < trewqasd \\leq lskdjfgh-1$ such that $bfhgtrde = cdrfvtgb$. Then\n\\[\nqzxwvtnp = asdfghjk^{ghjklpoi} zxcvbnml^{qazxswec} \\cdots asdfghjk^{asqwerty_{poiuylkj}} zxcvbnml^{zmbxcvnl_{poiuylkj}} asdfghjk^{asqwerty_{trewqasd+1}} zxcvbnml^{zmbxcvnl_{trewqasd+1}} \\cdots asdfghjk^{uiojklmn} zxcvbnml^{plmkoijn}\n\\]\nis another representation of $qzxwvtnp$ of length strictly less than $lskdjfgh$, a contradiction.\n\n\\noindent\n\\textbf{Remark:}\nIf one considers $tgbnhyuj,\\dots,wsxedcrf$ instead of $edcbgfra,\\dots,bfhgtrde$, then the case $qzxwvtnp=poiuytre$ must be handled separately: otherwise, one might end up with a representation of length 0 which is disallowed by the problem statement.\n\n\\noindent\n\\textbf{Reinterpretation:} \nNote that the elements $asdfghjk zxcvbnml, asdfghjk zxcvbnml^{-1}, asdfghjk^{-1} zxcvbnml, asdfghjk^{-1} zxcvbnml^{-1}$ generate $asdfghjk zxcvbnml (asdfghjk^{-1} zxcvbnml)^{-1} = asdfghjk^{2}$ and hence all of $qwertyop$ (again using the hypothesis that $asdfghjk$ has odd order, as above). Form the Cayley digraph on the set $qwertyop$, i.e., the directed graph with an edge from $tgbnhyuj$ to $s_2$ whenever $s_2 = tgbnhyuj *$ for $* \\in \\{asdfghjk zxcvbnml, asdfghjk zxcvbnml^{-1}, asdfghjk^{-1} zxcvbnml, asdfghjk^{-1} zxcvbnml^{-1}\\}$. Since $qwertyop$ is finite, this digraph is strongly connected: there exists at least one path from any vertex to any other vertex (traveling all edges in the correct direction). The shortest such path cannot repeat any vertices (except the starting and ending vertices in case they coincide), and so has length at most $|qwertyop|$.\n\n\\noindent\n\\textbf{Second solution:}\nFor $lskdjfgh$ a positive integer, let $nhygtvfr$ be the set of $qzxwvtnp \\in qwertyop$ which admit a representation of length at most $lskdjfgh$ (terminology as in the first solution); obviously $nhygtvfr \\subseteq nhygtvfr_{lskdjfgh+1}$. We will show that $nhygtvfr \\neq nhygtvfr_{lskdjfgh+1}$ unless $nhygtvfr = qwertyop$; this will imply by induction on $lskdjfgh$ that $\\#nhygtvfr \\geq \\min\\{lskdjfgh, |qwertyop|\\}$ and hence that $nhygtvfr = qwertyop$ for some $lskdjfgh \\leq |qwertyop|$.\n\nSuppose that $nhygtvfr = nhygtvfr_{lskdjfgh+1}$. Then the map $qzxwvtnp \\mapsto qzxwvtnp asdfghjk zxcvbnml$ defines an injective map $nhygtvfr \\to nhygtvfr_{lskdjfgh+1} = nhygtvfr$, so $nhygtvfr$ is closed under right multiplication by $asdfghjk zxcvbnml$. By the same token, $nhygtvfr$ is closed under right multiplication by each of $asdfghjk zxcvbnml^{-1}, asdfghjk^{-1} zxcvbnml, asdfghjk^{-1} zxcvbnml^{-1}$. Since these elements generate $qwertyop$ as in the first solution, it follows that $nhygtvfr = qwertyop$ as claimed.\n\n\\noindent\n\\textbf{Remark:} The condition on the order of $asdfghjk$ is needed to rule out the case where $qwertyop$ admits a (necessarily normal) subgroup $lkjhgfds$ of index 2 not containing either $asdfghjk$ or $zxcvbnml$; in this case, all products of the indicated form belong to $lkjhgfds$. On the other hand, if one assumes that both $asdfghjk$ and $zxcvbnml$ have odd order, then one can say a bit more: there exists some positive integer $lskdjfgh$ with $1 \\leq lskdjfgh \\leq |qwertyop|$ such that every element of $qwertyop$ has a representation of length exactly $lskdjfgh$. (Namely, the set of such elements for a given $lskdjfgh$ strictly increases in size until it is stable under right multiplication by both\n$asdfghjk zxcvbnml (asdfghjk^{-1} zxcvbnml)^{-1} = asdfghjk^{2}$ and $asdfghjk zxcvbnml (asdfghjk zxcvbnml^{-1})^{-1} = asdfghjk zxcvbnml^{2} asdfghjk^{-1}$, but under the present hypotheses these generate $qwertyop$.)"
    },
    "kernel_variant": {
      "question": "Let $G$ be a finite group generated by the two elements $g,h$ and suppose that $\\operatorname{ord}(g)$ is odd.  \nChoose two {\\em odd} positive integers  \n\\[\nP,\\;Q\\quad\\text{with}\\quad \n\\gcd\\!\\bigl(P,\\operatorname{ord}(g)\\bigr)=1, \n\\qquad \n\\gcd\\!\\bigl(Q,\\operatorname{ord}(h)\\bigr)=1 .\n\\tag{$\\ast$}\n\\]\n\nFor every {\\em strictly alternating} word  \n\\[\nw \\;=\\; g^{m_{1}}h^{\\,n_{1}}g^{m_{2}}h^{\\,n_{2}}\\dotsm g^{m_{r}}h^{\\,n_{r}},\n\\qquad m_{i},n_{i}\\in\\{-1,1\\},\n\\]\ndefine the $P$– and $Q$–signatures\n\\[\n\\sigma_{g}(w)=\\sum_{i=1}^{r} m_{i}\\pmod{P},\n\\qquad\n\\sigma_{h}(w)=\\sum_{i=1}^{r} n_{i}\\pmod{Q}.\n\\]\n\nThe word $w$ is called $(P,Q)$–{\\em balanced} if  \n\\[\n\\sigma_{g}(w)\\equiv0\\pmod{P},\n\\qquad\n\\sigma_{h}(w)\\equiv0\\pmod{Q}.\n\\]\n\nProve that for every element $s\\in G$ there exists a $(P,Q)$–balanced strictly alternating word $w$ representing $s$ whose pair–length $r$ satisfies\n\\[\n1\\;\\le\\; r\\;\\le\\; |G|\\,P\\,Q .\n\\]\n\nConsequently $w$ contains at most $2\\,|G|\\,P\\,Q$ individual letters.  \n(When $s=e$ the empty word is admissible, but the statement asserts that a {\\em non-empty} balanced word of length at most $|G|PQ$ can always be found.)\n\n\\bigskip",
      "solution": "We again enlarge the group, but we first repair the generation problem that arose in the review.\n\n\\medskip\n1.  The auxiliary group.  \nSet  \n\\[\n\\Gamma \\;=\\; G \\times \\mathbf Z_{P}\\times\\mathbf Z_{Q},\n\\qquad |\\Gamma| = |G|\\,P\\,Q .\n\\]\nElements of $\\Gamma$ are written multiplicatively, while the second and\nthird coordinates are added modulo $P$ and $Q$, respectively.\nIntroduce   \n\\[\n\\gamma \\;:=\\; (g,1,0),\\qquad\n\\eta   \\;:=\\; (h,0,1)\\;\\in\\Gamma .\n\\]\n\n\\medskip\n2.  $\\gamma,\\eta$ generate $\\Gamma$, and $\\gamma$ has odd order.  \n\n\\noindent\n(a)  {\\em Generation.}  \nBecause $g,h$ generate $G$, every $x\\in G$ can be written\n\\[\nx=g^{u_{1}}h^{v_{1}}g^{u_{2}}h^{v_{2}}\\dotsm g^{u_{k}}h^{v_{k}}\n\\qquad(u_{j},v_{j}\\in\\{-1,1\\}).\n\\tag{1}\n\\]\n\nFor $a\\in\\mathbf Z_{P}$ and $b\\in\\mathbf Z_{Q}$\nchoose representatives $\\tilde a,\\tilde b\\in\\mathbf Z$.\nUsing (1) we obtain\n\\[\n(x,\\tilde a,\\tilde b)\n\\;=\\;\n\\gamma^{u_{1}}\\eta^{v_{1}}\\gamma^{u_{2}}\\eta^{v_{2}}\n\\dotsm\\gamma^{u_{k}}\\eta^{v_{k}}\\,\n(\\,e,\\tilde a-\\Sigma_{g},\\tilde b-\\Sigma_{h}),\n\\]\nwhere $\\Sigma_{g}:=\\sum_{j}u_{j}$ and\n$\\Sigma_{h}:=\\sum_{j}v_{j}$.\nHence it suffices to show that\n\\[\n(e,1,0),\\;(e,0,1)\\;\\in\\langle\\gamma,\\eta\\rangle .\n\\tag{2}\n\\]\n\nSince $\\gcd\\!\\bigl(P,\\operatorname{ord}(g)\\bigr)=1$ there exist integers\n$\\alpha,\\beta$ with\n\\[\n\\alpha\\,\\operatorname{ord}(g)+\\beta\\,P=1 .\n\\]\nBecause $g^{\\operatorname{ord}(g)}=e$ we have\n\\[\n\\gamma^{\\alpha\\,\\operatorname{ord}(g)}\n= (e,\\alpha\\,\\operatorname{ord}(g),0)\n= (e,1,0).\n\\]\nThus $(e,1,0)$ belongs to $\\langle\\gamma,\\eta\\rangle$.\nAn entirely analogous argument, using\n$\\gcd\\!\\bigl(Q,\\operatorname{ord}(h)\\bigr)=1$, gives $(e,0,1)$.\nTherefore (2) holds, whence $\\langle\\gamma,\\eta\\rangle=\\Gamma$.\n\n\\smallskip\n(b)  {\\em Order of $\\gamma$.}  \nWrite $d:=\\operatorname{ord}(g)$ (an odd integer).\nSince $\\gcd(d,P)=1$ and both numbers are odd,\n\\[\n\\operatorname{ord}(\\gamma)=\\operatorname{lcm}(d,P),\n\\]\nwhich is again odd.  Hence $\\Gamma$ fulfils the hypotheses of\nthe {\\em original} theorem (finite, two generators, one of odd order).\n\n\\medskip\n3.  A short alternating representation in $\\Gamma$.  \nApply the original theorem to $\\Gamma$ and\n$\\overline s:=(s,0,0)\\in\\Gamma$.  We obtain a strictly alternating word\n\\[\n\\widetilde w\n=\\gamma^{m_{1}}\\eta^{\\,n_{1}}\\gamma^{m_{2}}\\eta^{\\,n_{2}}\n\\dotsm\\gamma^{m_{r}}\\eta^{\\,n_{r}},\n\\qquad m_{i},n_{i}\\in\\{-1,1\\},\n\\]\nsatisfying\n\\[\n\\widetilde w=\\overline s,\n\\qquad\n1\\le r\\le |\\Gamma|=|G|\\,P\\,Q .\n\\tag{3}\n\\]\n\n\\medskip\n4.  Projection to $G$ and the two signatures.  \nProjecting the word $\\widetilde w$ to the first coordinate gives\n\\[\nw \\;=\\; g^{m_{1}}h^{\\,n_{1}}g^{m_{2}}h^{\\,n_{2}}\\dotsm g^{m_{r}}h^{\\,n_{r}}\n\\in G,\n\\]\nand equality $\\widetilde w=\\overline s$ yields $w=s$.\n\nThe second coordinate of $\\widetilde w$ is $0\\in\\mathbf Z_{P}$, hence\n\\[\n0=\\sum_{i=1}^{r} m_{i} = \\sigma_{g}(w)\\pmod{P}.\n\\]\nLikewise the third coordinate being $0\\in\\mathbf Z_{Q}$ gives\n$\\sigma_{h}(w)\\equiv0\\pmod{Q}$.\nThus $w$ is $(P,Q)$–balanced.\n\n\\medskip\n5.  Ensuring non-emptiness when $s=e$.  \nIf $s\\neq e$, the word $w$ is non-empty by (3).\nIf $s=e$ and $r=0$ were to occur, replace $w$ by\n\\[\ng\\,h\\,g^{-1}h^{-1},\n\\]\nwhich is strictly alternating, $(P,Q)$–balanced, and has\npair-length $2\\le |G|PQ$.\nHence\n\\[\n1\\le r\\le|G|PQ\n\\]\nalways holds.\n\n\\medskip\n6.  Letter count.  \nEach pair contributes two letters, so the total number of letters is\n$\\le 2\\,|G|\\,P\\,Q$.\n\n\\hfill$\\square$\n\n\\bigskip",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.842691",
        "was_fixed": false,
        "difficulty_analysis": "1.  Additional structure:  The solver must work not only in G but in the product Γ=G×ℤ_P×ℤ_Q, thus simultaneously tracking the group element and two independent residue conditions.\n2.  New constraints:  Representations must satisfy two congruence equations (σ_g≡0 mod P, σ_h≡0 mod Q) in addition to the alternating ±1 pattern.\n3.  Heavier machinery:  The proof employs Cayley-graph connectivity of a product group and BFS–style diameter arguments rather than the elementary pigeonhole principle on G alone.\n4.  Coprimality and oddness enter twice: to ensure that the four step–vectors generate the whole residue group and that the original odd-order argument still lets g^{±1} and h^{±1} steer through Γ.\n5.  Quantitatively larger bound:  The upper bound |G|·P·Q grows with the new ambient group, reflecting the increased combinatorial complexity.\n\nThese new layers of bookkeeping, coördinate tracking, and graph-theoretic reasoning render the enhanced variant substantially harder than both the original exercise and the current kernel version."
      }
    },
    "original_kernel_variant": {
      "question": "Let $G$ be a finite group generated by the two elements $g,h$ and suppose that $\\operatorname{ord}(g)$ is odd.  \nChoose two {\\em odd} positive integers  \n\\[\nP,\\;Q\\quad\\text{with}\\quad \n\\gcd\\!\\bigl(P,\\operatorname{ord}(g)\\bigr)=1, \n\\qquad \n\\gcd\\!\\bigl(Q,\\operatorname{ord}(h)\\bigr)=1 .\n\\tag{$\\ast$}\n\\]\n\nFor every {\\em strictly alternating} word  \n\\[\nw \\;=\\; g^{m_{1}}h^{\\,n_{1}}g^{m_{2}}h^{\\,n_{2}}\\dotsm g^{m_{r}}h^{\\,n_{r}},\n\\qquad m_{i},n_{i}\\in\\{-1,1\\},\n\\]\ndefine the $P$– and $Q$–signatures\n\\[\n\\sigma_{g}(w)=\\sum_{i=1}^{r} m_{i}\\pmod{P},\n\\qquad\n\\sigma_{h}(w)=\\sum_{i=1}^{r} n_{i}\\pmod{Q}.\n\\]\n\nThe word $w$ is called $(P,Q)$–{\\em balanced} if  \n\\[\n\\sigma_{g}(w)\\equiv0\\pmod{P},\n\\qquad\n\\sigma_{h}(w)\\equiv0\\pmod{Q}.\n\\]\n\nProve that for every element $s\\in G$ there exists a $(P,Q)$–balanced strictly alternating word $w$ representing $s$ whose pair–length $r$ satisfies\n\\[\n1\\;\\le\\; r\\;\\le\\; |G|\\,P\\,Q .\n\\]\n\nConsequently $w$ contains at most $2\\,|G|\\,P\\,Q$ individual letters.  \n(When $s=e$ the empty word is admissible, but the statement asserts that a {\\em non-empty} balanced word of length at most $|G|PQ$ can always be found.)\n\n\\bigskip",
      "solution": "We again enlarge the group, but we first repair the generation problem that arose in the review.\n\n\\medskip\n1.  The auxiliary group.  \nSet  \n\\[\n\\Gamma \\;=\\; G \\times \\mathbf Z_{P}\\times\\mathbf Z_{Q},\n\\qquad |\\Gamma| = |G|\\,P\\,Q .\n\\]\nElements of $\\Gamma$ are written multiplicatively, while the second and\nthird coordinates are added modulo $P$ and $Q$, respectively.\nIntroduce   \n\\[\n\\gamma \\;:=\\; (g,1,0),\\qquad\n\\eta   \\;:=\\; (h,0,1)\\;\\in\\Gamma .\n\\]\n\n\\medskip\n2.  $\\gamma,\\eta$ generate $\\Gamma$, and $\\gamma$ has odd order.  \n\n\\noindent\n(a)  {\\em Generation.}  \nBecause $g,h$ generate $G$, every $x\\in G$ can be written\n\\[\nx=g^{u_{1}}h^{v_{1}}g^{u_{2}}h^{v_{2}}\\dotsm g^{u_{k}}h^{v_{k}}\n\\qquad(u_{j},v_{j}\\in\\{-1,1\\}).\n\\tag{1}\n\\]\n\nFor $a\\in\\mathbf Z_{P}$ and $b\\in\\mathbf Z_{Q}$\nchoose representatives $\\tilde a,\\tilde b\\in\\mathbf Z$.\nUsing (1) we obtain\n\\[\n(x,\\tilde a,\\tilde b)\n\\;=\\;\n\\gamma^{u_{1}}\\eta^{v_{1}}\\gamma^{u_{2}}\\eta^{v_{2}}\n\\dotsm\\gamma^{u_{k}}\\eta^{v_{k}}\\,\n(\\,e,\\tilde a-\\Sigma_{g},\\tilde b-\\Sigma_{h}),\n\\]\nwhere $\\Sigma_{g}:=\\sum_{j}u_{j}$ and\n$\\Sigma_{h}:=\\sum_{j}v_{j}$.\nHence it suffices to show that\n\\[\n(e,1,0),\\;(e,0,1)\\;\\in\\langle\\gamma,\\eta\\rangle .\n\\tag{2}\n\\]\n\nSince $\\gcd\\!\\bigl(P,\\operatorname{ord}(g)\\bigr)=1$ there exist integers\n$\\alpha,\\beta$ with\n\\[\n\\alpha\\,\\operatorname{ord}(g)+\\beta\\,P=1 .\n\\]\nBecause $g^{\\operatorname{ord}(g)}=e$ we have\n\\[\n\\gamma^{\\alpha\\,\\operatorname{ord}(g)}\n= (e,\\alpha\\,\\operatorname{ord}(g),0)\n= (e,1,0).\n\\]\nThus $(e,1,0)$ belongs to $\\langle\\gamma,\\eta\\rangle$.\nAn entirely analogous argument, using\n$\\gcd\\!\\bigl(Q,\\operatorname{ord}(h)\\bigr)=1$, gives $(e,0,1)$.\nTherefore (2) holds, whence $\\langle\\gamma,\\eta\\rangle=\\Gamma$.\n\n\\smallskip\n(b)  {\\em Order of $\\gamma$.}  \nWrite $d:=\\operatorname{ord}(g)$ (an odd integer).\nSince $\\gcd(d,P)=1$ and both numbers are odd,\n\\[\n\\operatorname{ord}(\\gamma)=\\operatorname{lcm}(d,P),\n\\]\nwhich is again odd.  Hence $\\Gamma$ fulfils the hypotheses of\nthe {\\em original} theorem (finite, two generators, one of odd order).\n\n\\medskip\n3.  A short alternating representation in $\\Gamma$.  \nApply the original theorem to $\\Gamma$ and\n$\\overline s:=(s,0,0)\\in\\Gamma$.  We obtain a strictly alternating word\n\\[\n\\widetilde w\n=\\gamma^{m_{1}}\\eta^{\\,n_{1}}\\gamma^{m_{2}}\\eta^{\\,n_{2}}\n\\dotsm\\gamma^{m_{r}}\\eta^{\\,n_{r}},\n\\qquad m_{i},n_{i}\\in\\{-1,1\\},\n\\]\nsatisfying\n\\[\n\\widetilde w=\\overline s,\n\\qquad\n1\\le r\\le |\\Gamma|=|G|\\,P\\,Q .\n\\tag{3}\n\\]\n\n\\medskip\n4.  Projection to $G$ and the two signatures.  \nProjecting the word $\\widetilde w$ to the first coordinate gives\n\\[\nw \\;=\\; g^{m_{1}}h^{\\,n_{1}}g^{m_{2}}h^{\\,n_{2}}\\dotsm g^{m_{r}}h^{\\,n_{r}}\n\\in G,\n\\]\nand equality $\\widetilde w=\\overline s$ yields $w=s$.\n\nThe second coordinate of $\\widetilde w$ is $0\\in\\mathbf Z_{P}$, hence\n\\[\n0=\\sum_{i=1}^{r} m_{i} = \\sigma_{g}(w)\\pmod{P}.\n\\]\nLikewise the third coordinate being $0\\in\\mathbf Z_{Q}$ gives\n$\\sigma_{h}(w)\\equiv0\\pmod{Q}$.\nThus $w$ is $(P,Q)$–balanced.\n\n\\medskip\n5.  Ensuring non-emptiness when $s=e$.  \nIf $s\\neq e$, the word $w$ is non-empty by (3).\nIf $s=e$ and $r=0$ were to occur, replace $w$ by\n\\[\ng\\,h\\,g^{-1}h^{-1},\n\\]\nwhich is strictly alternating, $(P,Q)$–balanced, and has\npair-length $2\\le |G|PQ$.\nHence\n\\[\n1\\le r\\le|G|PQ\n\\]\nalways holds.\n\n\\medskip\n6.  Letter count.  \nEach pair contributes two letters, so the total number of letters is\n$\\le 2\\,|G|\\,P\\,Q$.\n\n\\hfill$\\square$\n\n\\bigskip",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.645041",
        "was_fixed": false,
        "difficulty_analysis": "1.  Additional structure:  The solver must work not only in G but in the product Γ=G×ℤ_P×ℤ_Q, thus simultaneously tracking the group element and two independent residue conditions.\n2.  New constraints:  Representations must satisfy two congruence equations (σ_g≡0 mod P, σ_h≡0 mod Q) in addition to the alternating ±1 pattern.\n3.  Heavier machinery:  The proof employs Cayley-graph connectivity of a product group and BFS–style diameter arguments rather than the elementary pigeonhole principle on G alone.\n4.  Coprimality and oddness enter twice: to ensure that the four step–vectors generate the whole residue group and that the original odd-order argument still lets g^{±1} and h^{±1} steer through Γ.\n5.  Quantitatively larger bound:  The upper bound |G|·P·Q grows with the new ambient group, reflecting the increased combinatorial complexity.\n\nThese new layers of bookkeeping, coördinate tracking, and graph-theoretic reasoning render the enhanced variant substantially harder than both the original exercise and the current kernel version."
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}