{
  "index": "2007-A-5",
  "type": "NT",
  "tag": [
    "NT",
    "COMB"
  ],
  "difficulty": "",
  "question": "Suppose that a finite group has exactly $n$ elements of order $p$,\nwhere $p$ is a prime. Prove that either $n=0$ or $p$ divides $n+1$.",
  "solution": "In all solutions, let $G$ be a finite group of order $m$.\n\n\\textbf{First solution:}\nBy Lagrange's theorem, if $m$ is\nnot divisible by $p$, then $n = 0$. Otherwise, let $S$ be the set of\n$p$-tuples $(a_0,\\dots,a_{p-1}) \\in G^p$ such that $a_0 \\cdots a_{p-1} = e$;\nthen $S$ has cardinality $m^{p-1}$, which is divisible by $p$.\nNote that this set is invariant under cyclic permutation, that is,\nif $(a_0,\\dots,a_{p-1}) \\in S$, then $(a_1,\\dots,a_{p-1},a_0) \\in S$ also.\nThe fixed points under this operation are the tuples\n$(a,\\dots,a)$ with $a^p = e$; all other tuples can be grouped into orbits\nunder cyclic permutation, each of which has size $p$. Consequently,\nthe number of $a \\in G$ with $a^p = e$ is divisible by $p$; since that number\nis $n+1$ (only $e$ has order 1), this proves the claim.\n\n\\textbf{Second solution:}\n(by Anand Deopurkar)\nAssume that $n > 0$, and let $H$ be any subgroup of $G$ of order $p$.\nLet $S$ be the set of all elements of $G \\setminus H$\nof order dividing $p$, and let\n$H$ act on $G$ by conjugation. Each orbit has size $p$ except for those\nwhich consist of individual elements $g$ which commute with $H$.\nFor each such $g$, $g$ and $H$ generate an elementary abelian subgroup of\n$G$ of order $p^2$. However, we can group these $g$ into sets of size\n$p^2-p$ based on which subgroup they generate together with $H$.\nHence the cardinality of $S$ is divisible by $p$; adding the $p-1$\nnontrivial elements of $H$ gives $n \\equiv -1 \\pmod{p}$ as desired.\n\n\\textbf{Third solution:}\nLet $S$ be the set of elements in $G$ having order dividing $p$, and let\n$H$ be an elementary abelian $p$-group of maximal order in $G$.  If\n$|H|=1$, then we are done.  So assume $|H|=p^k$ for some $k\\geq 1$, and\nlet $H$ act on $S$ by conjugation.  Let $T\\subset S$ denote the set of\nfixed points of this action.  Then the size of every $H$-orbit on $S$\ndivides $p^k$, and so $|S|\\equiv |T| \\pmod{p}$.\nOn the other hand, $H\\subset T$, and if $T$ contained an element not in $H$,\nthen that would contradict the maximality of $H$.  It follows that\n$H=T$, and so $|S|\\equiv |T| = |H| = p^k \\equiv 0 \\pmod{p}$,\ni.e., $|S|=n+1$ is a multiple of $p$.\n\n\\textbf{Remark:} This result is a theorem of Cauchy; the first solution\nabove is due to McKay. A more general (and more difficult) result was proved\nby Frobenius: for any positive integer $m$, if $G$ is a finite group of order\ndivisible by $m$, then the number of elements of $G$ of order dividing $m$\nis a multiple of $m$.",
  "vars": [
    "G",
    "S",
    "H",
    "T",
    "g",
    "a",
    "a_0",
    "a_1",
    "a_p-1",
    "e"
  ],
  "params": [
    "n",
    "p",
    "m",
    "k"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "G": "finitegroup",
        "S": "orderset",
        "H": "psubgroup",
        "T": "fixedset",
        "g": "groupelem",
        "a": "tupleelem",
        "a_0": "tuplezero",
        "a_1": "tupleone",
        "a_{p-1}": "tuplelast",
        "n": "ordercount",
        "p": "primeorder",
        "m": "grouporder",
        "k": "rankpower"
      },
      "question": "Suppose that a finite group has exactly $ordercount$ elements of order $primeorder$, where $primeorder$ is a prime. Prove that either $ordercount=0$ or $primeorder$ divides $ordercount+1$.",
      "solution": "In all solutions, let $finitegroup$ be a finite group of order $grouporder$.\n\n\\textbf{First solution:}\nBy Lagrange's theorem, if $grouporder$ is\nnot divisible by $primeorder$, then $ordercount = 0$. Otherwise, let $orderset$ be the set of\n$primeorder$-tuples $(tuplezero,\\dots,tuplelast) \\in finitegroup^{primeorder}$ such that $tuplezero \\cdots tuplelast = e$;\nthen $orderset$ has cardinality $grouporder^{primeorder-1}$, which is divisible by $primeorder$.\nNote that this set is invariant under cyclic permutation, that is,\nif $(tuplezero,\\dots,tuplelast) \\in orderset$, then $(tupleone,\\dots,tuplelast,tuplezero) \\in orderset$ also.\nThe fixed points under this operation are the tuples\n$(tupleelem,\\dots,tupleelem)$ with $tupleelem^{primeorder} = e$; all other tuples can be grouped into orbits\nunder cyclic permutation, each of which has size $primeorder$. Consequently,\nthe number of $tupleelem \\in finitegroup$ with $tupleelem^{primeorder} = e$ is divisible by $primeorder$; since that number\nis $ordercount+1$ (only $e$ has order 1), this proves the claim.\n\n\\textbf{Second solution:}\n(by Anand Deopurkar)\nAssume that $ordercount > 0$, and let $psubgroup$ be any subgroup of $finitegroup$ of order $primeorder$.\nLet $orderset$ be the set of all elements of $finitegroup \\setminus psubgroup$\nof order dividing $primeorder$, and let\n$psubgroup$ act on $finitegroup$ by conjugation. Each orbit has size $primeorder$ except for those\nwhich consist of individual elements $groupelem$ which commute with $psubgroup$.\nFor each such $groupelem$, $groupelem$ and $psubgroup$ generate an elementary abelian subgroup of\n$finitegroup$ of order $primeorder^2$. However, we can group these $groupelem$ into sets of size\n$primeorder^2-primeorder$ based on which subgroup they generate together with $psubgroup$.\nHence the cardinality of $orderset$ is divisible by $primeorder$; adding the $primeorder-1$\nnontrivial elements of $psubgroup$ gives $ordercount \\equiv -1 \\pmod{primeorder}$ as desired.\n\n\\textbf{Third solution:}\nLet $orderset$ be the set of elements in $finitegroup$ having order dividing $primeorder$, and let\n$psubgroup$ be an elementary abelian $primeorder$-group of maximal order in $finitegroup$.  If\n$|psubgroup|=1$, then we are done.  So assume $|psubgroup|=primeorder^{rankpower}$ for some $rankpower\\geq 1$, and\nlet $psubgroup$ act on $orderset$ by conjugation.  Let $fixedset\\subset orderset$ denote the set of\nfixed points of this action.  Then the size of every $psubgroup$-orbit on $orderset$\ndivides $primeorder^{rankpower}$, and so $|orderset|\\equiv |fixedset| \\pmod{primeorder}$.\nOn the other hand, $psubgroup\\subset fixedset$, and if $fixedset$ contained an element not in $psubgroup$,\nthen that would contradict the maximality of $psubgroup$.  It follows that\n$psubgroup=fixedset$, and so $|orderset|\\equiv |fixedset| = |psubgroup| = primeorder^{rankpower} \\equiv 0 \\pmod{primeorder}$,\ni.e., $|orderset|=ordercount+1$ is a multiple of $primeorder$.\n\n\\textbf{Remark:} This result is a theorem of Cauchy; the first solution\nabove is due to McKay. A more general (and more difficult) result was proved\nby Frobenius: for any positive integer $grouporder$, if $finitegroup$ is a finite group of order\ndivisible by $grouporder$, then the number of elements of $finitegroup$ of order dividing $grouporder$\nis a multiple of $grouporder$."
    },
    "descriptive_long_confusing": {
      "map": {
        "G": "labyrinth",
        "S": "turnpike",
        "H": "horsepower",
        "T": "timeshift",
        "g": "stoneware",
        "a": "driftwood",
        "a_0": "driftwoodzero",
        "a_1": "driftwoodone",
        "a_{p-1}": "driftwoodlast",
        "n": "wavelength",
        "p": "raincloud",
        "m": "undertone",
        "k": "slingback"
      },
      "question": "Suppose that a finite group has exactly $wavelength$ elements of order $raincloud$,\nwhere $raincloud$ is a prime. Prove that either $wavelength=0$ or $raincloud$ divides $wavelength+1$.",
      "solution": "In all solutions, let $labyrinth$ be a finite group of order $undertone$.\n\n\\textbf{First solution:}\nBy Lagrange's theorem, if $undertone$ is\nnot divisible by $raincloud$, then $wavelength = 0$. Otherwise, let $turnpike$ be the set of\n$raincloud$-tuples $(driftwoodzero,\\dots,driftwoodlast) \\in labyrinth^{raincloud}$ such that $driftwoodzero \\cdots driftwoodlast = e$;\nthen $turnpike$ has cardinality $undertone^{raincloud-1}$, which is divisible by $raincloud$.\nNote that this set is invariant under cyclic permutation, that is,\nif $(driftwoodzero,\\dots,driftwoodlast) \\in turnpike$, then $(driftwoodone,\\dots,driftwoodlast,driftwoodzero) \\in turnpike$ also.\nThe fixed points under this operation are the tuples\n$(driftwood,\\dots,driftwood)$ with $driftwood^{raincloud} = e$; all other tuples can be grouped into orbits\nunder cyclic permutation, each of which has size $raincloud$. Consequently,\nthe number of $driftwood \\in labyrinth$ with $driftwood^{raincloud} = e$ is divisible by $raincloud$; since that number\nis $wavelength+1$ (only $e$ has order 1), this proves the claim.\n\n\\textbf{Second solution:}\n(by Anand Deopurkar)\nAssume that $wavelength > 0$, and let $horsepower$ be any subgroup of $labyrinth$ of order $raincloud$.\nLet $turnpike$ be the set of all elements of $labyrinth \\setminus horsepower$\nof order dividing $raincloud$, and let\n$horsepower$ act on $labyrinth$ by conjugation. Each orbit has size $raincloud$ except for those\nwhich consist of individual elements $stoneware$ which commute with $horsepower$.\nFor each such $stoneware$, $stoneware$ and $horsepower$ generate an elementary abelian subgroup of\n$labyrinth$ of order $raincloud^{2}$. However, we can group these $stoneware$ into sets of size\n$raincloud^{2}-raincloud$ based on which subgroup they generate together with $horsepower$.\nHence the cardinality of $turnpike$ is divisible by $raincloud$; adding the $raincloud-1$\nnontrivial elements of $horsepower$ gives $wavelength \\equiv -1 \\pmod{raincloud}$ as desired.\n\n\\textbf{Third solution:}\nLet $turnpike$ be the set of elements in $labyrinth$ having order dividing $raincloud$, and let\n$horsepower$ be an elementary abelian $raincloud$-group of maximal order in $labyrinth$.  If\n$|horsepower|=1$, then we are done.  So assume $|horsepower|=raincloud^{slingback}$ for some $slingback\\geq 1$, and\nlet $horsepower$ act on $turnpike$ by conjugation.  Let $timeshift\\subset turnpike$ denote the set of\nfixed points of this action.  Then the size of every $horsepower$-orbit on $turnpike$\ndivides $raincloud^{slingback}$, and so $|turnpike|\\equiv |timeshift| \\pmod{raincloud}$.\nOn the other hand, $horsepower\\subset timeshift$, and if $timeshift$ contained an element not in $horsepower$,\nthen that would contradict the maximality of $horsepower$.  It follows that\n$horsepower=timeshift$, and so $|turnpike|\\equiv |timeshift| = |horsepower| = raincloud^{slingback} \\equiv 0 \\pmod{raincloud}$,\ni.e., $|turnpike|=wavelength+1$ is a multiple of $raincloud$.\n\n\\textbf{Remark:} This result is a theorem of Cauchy; the first solution\nabove is due to McKay. A more general (and more difficult) result was proved\nby Frobenius: for any positive integer $undertone$, if $labyrinth$ is a finite group of order\ndivisible by $undertone$, then the number of elements of $labyrinth$ of order dividing $undertone$\nis a multiple of $undertone$.}"
    },
    "descriptive_long_misleading": {
      "map": {
        "G": "chaosset",
        "S": "emptiness",
        "H": "outsider",
        "T": "wanderers",
        "g": "aggregate",
        "a": "absence",
        "a_0": "ultimate",
        "a_1": "terminal",
        "a_p-1": "beginner",
        "e": "opposite",
        "n": "uncounted",
        "p": "composite",
        "m": "fraction",
        "k": "rootless"
      },
      "question": "Suppose that a finite group has exactly $uncounted$ elements of order $composite$,\nwhere $composite$ is a prime. Prove that either $uncounted=0$ or $composite$ divides $uncounted+1$.",
      "solution": "In all solutions, let $chaosset$ be a finite group of order $fraction$.\n\n\\textbf{First solution:}\nBy Lagrange's theorem, if $fraction$ is\nnot divisible by $composite$, then $uncounted = 0$. Otherwise, let $emptiness$ be the set of\n$composite$-tuples $(ultimate,\\dots,beginner) \\in chaosset^{composite}$ such that $ultimate \\cdots beginner = opposite$;\nthen $emptiness$ has cardinality $fraction^{composite-1}$, which is divisible by $composite$.\nNote that this set is invariant under cyclic permutation, that is,\nif $(ultimate,\\dots,beginner) \\in emptiness$, then $(terminal,\\dots,beginner,ultimate) \\in emptiness$ also.\nThe fixed points under this operation are the tuples\n$(absence,\\dots,absence)$ with $absence^{composite} = opposite$; all other tuples can be grouped into orbits\nunder cyclic permutation, each of which has size $composite$. Consequently,\nthe number of $absence \\in chaosset$ with $absence^{composite} = opposite$ is divisible by $composite$; since that number\nis $uncounted+1$ (only $opposite$ has order 1), this proves the claim.\n\n\\textbf{Second solution:}\n(by Anand Deopurkar)\nAssume that $uncounted > 0$, and let $outsider$ be any subgroup of $chaosset$ of order $composite$.\nLet $emptiness$ be the set of all elements of $chaosset \\setminus outsider$\nof order dividing $composite$, and let\n$outsider$ act on $chaosset$ by conjugation. Each orbit has size $composite$ except for those\nwhich consist of individual elements $aggregate$ which commute with $outsider$.\nFor each such $aggregate$, $aggregate$ and $outsider$ generate an elementary abelian subgroup of\n$chaosset$ of order $composite^2$. However, we can group these $aggregate$ into sets of size\n$composite^2-composite$ based on which subgroup they generate together with $outsider$.\nHence the cardinality of $emptiness$ is divisible by $composite$; adding the $composite-1$\nnontrivial elements of $outsider$ gives $uncounted \\equiv -1 \\pmod{composite}$ as desired.\n\n\\textbf{Third solution:}\nLet $emptiness$ be the set of elements in $chaosset$ having order dividing $composite$, and let\n$outsider$ be an elementary abelian $composite$-group of maximal order in $chaosset$.  If\n$|outsider|=1$, then we are done.  So assume $|outsider|=composite^{rootless}$ for some $rootless\\geq 1$, and\nlet $outsider$ act on $emptiness$ by conjugation.  Let $wanderers\\subset emptiness$ denote the set of\nfixed points of this action.  Then the size of every $outsider$-orbit on $emptiness$\ndivides $composite^{rootless}$, and so $|emptiness|\\equiv |wanderers| \\pmod{composite}$.\nOn the other hand, $outsider\\subset wanderers$, and if $wanderers$ contained an element not in $outsider$,\nthen that would contradict the maximality of $outsider$.  It follows that\n$outsider=wanderers$, and so $|emptiness|\\equiv |wanderers| = |outsider| = composite^{rootless} \\equiv 0 \\pmod{composite}$,\ni.e., $|emptiness|=uncounted+1$ is a multiple of $composite$.\n\n\\textbf{Remark:} This result is a theorem of Cauchy; the first solution\nabove is due to McKay. A more general (and more difficult) result was proved\nby Frobenius: for any positive integer $fraction$, if $chaosset$ is a finite group of order\ndivisible by $fraction$, then the number of elements of $chaosset$ of order dividing $fraction$\nis a multiple of $fraction$.}",
      "confidence": 0.13
    },
    "garbled_string": {
      "map": {
        "G": "jzxqvtrm",
        "S": "vnglksad",
        "H": "pyhwmnqe",
        "T": "ldbxrsoc",
        "g": "tqwmslpa",
        "a": "yvzrnkie",
        "a_0": "yagldvop",
        "a_1": "nqpwlsae",
        "a_p-1": "ksrpvoae",
        "n": "hpqlsvzx",
        "p": "xdrmquab",
        "m": "flzvqust",
        "k": "pnahwiole"
      },
      "question": "Suppose that a finite group has exactly $hpqlsvzx$ elements of order $xdrmquab$,\nwhere $xdrmquab$ is a prime. Prove that either $hpqlsvzx=0$ or $xdrmquab$ divides $hpqlsvzx+1$.",
      "solution": "In all solutions, let $jzxqvtrm$ be a finite group of order $flzvqust$.\n\n\\textbf{First solution:}\nBy Lagrange's theorem, if $flzvqust$ is\nnot divisible by $xdrmquab$, then $hpqlsvzx = 0$. Otherwise, let $vnglksad$ be the set of\n$xdrmquab$-tuples $(yagldvop,\\dots,ksrpvoae) \\in jzxqvtrm^{xdrmquab}$ such that $yagldvop \\cdots ksrpvoae = e$;\nthen $vnglksad$ has cardinality $flzvqust^{xdrmquab-1}$, which is divisible by $xdrmquab$.\nNote that this set is invariant under cyclic permutation, that is,\nif $(yagldvop,\\dots,ksrpvoae) \\in vnglksad$, then $(nqpwlsae,\\dots,ksrpvoae,yagldvop) \\in vnglksad$ also.\nThe fixed points under this operation are the tuples\n$(yvzrnkie,\\dots,yvzrnkie)$ with $yvzrnkie^{xdrmquab} = e$; all other tuples can be grouped into orbits\nunder cyclic permutation, each of which has size $xdrmquab$. Consequently,\nthe number of $yvzrnkie \\in jzxqvtrm$ with $yvzrnkie^{xdrmquab} = e$ is divisible by $xdrmquab$; since that number\nis $hpqlsvzx+1$ (only $e$ has order 1), this proves the claim.\n\n\\textbf{Second solution:}\n(by Anand Deopurkar)\nAssume that $hpqlsvzx > 0$, and let $pyhwmnqe$ be any subgroup of $jzxqvtrm$ of order $xdrmquab$.\nLet $vnglksad$ be the set of all elements of $jzxqvtrm \\setminus pyhwmnqe$\nof order dividing $xdrmquab$, and let\n$pyhwmnqe$ act on $jzxqvtrm$ by conjugation. Each orbit has size $xdrmquab$ except for those\nwhich consist of individual elements $tqwmslpa$ which commute with $pyhwmnqe$.\nFor each such $tqwmslpa$, $tqwmslpa$ and $pyhwmnqe$ generate an elementary abelian subgroup of\n$jzxqvtrm$ of order $xdrmquab^2$. However, we can group these $tqwmslpa$ into sets of size\n$xdrmquab^2-xdrmquab$ based on which subgroup they generate together with $pyhwmnqe$.\nHence the cardinality of $vnglksad$ is divisible by $xdrmquab$; adding the $xdrmquab-1$\nnontrivial elements of $pyhwmnqe$ gives $hpqlsvzx \\equiv -1 \\pmod{xdrmquab}$ as desired.\n\n\\textbf{Third solution:}\nLet $vnglksad$ be the set of elements in $jzxqvtrm$ having order dividing $xdrmquab$, and let\n$pyhwmnqe$ be an elementary abelian $xdrmquab$-group of maximal order in $jzxqvtrm$.  If\n$|pyhwmnqe|=1$, then we are done.  So assume $|pyhwmnqe|=xdrmquab^{pnahwiole}$ for some $pnahwiole\\geq 1$, and\nlet $pyhwmnqe$ act on $vnglksad$ by conjugation.  Let $ldbxrsoc\\subset vnglksad$ denote the set of\nfixed points of this action.  Then the size of every $pyhwmnqe$-orbit on $vnglksad$\ndivides $xdrmquab^{pnahwiole}$, and so $|vnglksad|\\equiv |ldbxrsoc| \\pmod{xdrmquab}$.\nOn the other hand, $pyhwmnqe\\subset ldbxrsoc$, and if $ldbxrsoc$ contained an element not in $pyhwmnqe$,\nthen that would contradict the maximality of $pyhwmnqe$.  It follows that\n$pyhwmnqe=ldbxrsoc$, and so $|vnglksad|\\equiv |ldbxrsoc| = |pyhwmnqe| = xdrmquab^{pnahwiole} \\equiv 0 \\pmod{xdrmquab}$,\ni.e., $|vnglksad|=hpqlsvzx+1$ is a multiple of $xdrmquab$.\n\n\\textbf{Remark:} This result is a theorem of Cauchy; the first solution\nabove is due to McKay. A more general (and more difficult) result was proved\nby Frobenius: for any positive integer $flzvqust$, if $jzxqvtrm$ is a finite group of order\ndivisible by $flzvqust$, then the number of elements of $jzxqvtrm$ of order dividing $flzvqust$\nis a multiple of $flzvqust$.}",
      "confidence": "0.13"
    },
    "kernel_variant": {
      "question": "Let ℓ be a prime and r ≥ 1 an integer.  \nFor a finite group G set  \n\n C_r(G)= { (g₁,…,g_r)∈G^r  |  g_i^ℓ = e for every i and g_i g_j = g_j g_i for all i,j }.  \n\nThus C_r(G) consists of the ordered r–tuples of pairwise–commuting ℓ-elements of G.\n\nA)  Assume that ℓ divides |G|.  Prove that |C_r(G)| is divisible by ℓ.  \n   (If ℓ∤|G| then C_r(G)= {(e,…,e)}, so no divisibility can be expected.)\n\nB)  Show that the exponent obtained in A) is best possible:  \n\n for every prime ℓ and every r ≥ 1 there exists a finite group G with  \n  ν_ℓ(|C_r(G)|)=1.\n\nA single, uniform proof is expected; avoid a long case–by–case discussion.  \n(Hints for B:  for ℓ=2 one may take the symmetric group S₃; for odd ℓ one may take the group PSL₂(ℓ).)",
      "solution": "Throughout let ℓ be a fixed prime, r ≥ 1 an integer, and G a finite group.\n\n————————————————————————\nPart A.  Divisibility of |C_r(G)| by ℓ (under ℓ | |G|)  \nWe argue by induction on r.\n\nBase step r = 1.  \nPut  \n\n T := {(x₀,…,x_{ℓ-1}) ∈ G^{ℓ}  |  x₀x₁···x_{ℓ-1}=e}.  \n\nClearly |T| = |G|^{ℓ-1}, which is divisible by ℓ because ℓ | |G|.  \nLet the cyclic group C_ℓ = ⟨σ⟩ act on T by the left shift  \n σ·(x₀,…,x_{ℓ-1}) = (x₁,…,x_{ℓ-1},x₀).  \nAll orbits have size ℓ except the fixed points, which satisfy x₀ = ··· = x_{ℓ-1} and hence are precisely the diagonal tuples (x,…,x) with x^ℓ = e.  The class equation gives  \n\n |T| ≡ # {x∈G | x^ℓ=e}  (mod ℓ).  \n\nBecause |T| is a multiple of ℓ, the set S₁(G):= {x∈G | x^ℓ = e} has cardinality divisible by ℓ.  Now  \n\n |C₁(G)| = |S₁(G)| = ℓ·k     for some k∈ℕ,\n\nso |C₁(G)| is a multiple of ℓ, establishing the base case.\n\nInduction step.  Assume r ≥ 2 and that the statement holds for r−1.  \nFor h=(h₁,…,h_{r−1})∈C_{r−1}(G) put H:=⟨h₁,…,h_{r−1}⟩; this is an elementary abelian ℓ-subgroup, hence H ≤ C_G(H).  Define  \n\n S_h := { g∈C_G(H) | g^ℓ=e }.        (1)\n\nApply the already–proved base case r = 1 to the finite group C_G(H): because ℓ divides |C_G(H)| (H is non-trivial unless h=(e,…,e), but the argument below also works for the trivial H), the set S_h has size divisible by ℓ.  Next observe that an r-tuple in C_r(G) is obtained by adjoining to h any element of S_h as its last coordinate, and every such choice is admissible.  Hence the map  \n\n C_{r−1}(G) × S_h  →  C_r(G),    (h,g) ↦ (h₁,…,h_{r−1},g)\n\nis bijective inside each fibre over h, yielding  \n\n |C_r(G)| = ∑_{h∈C_{r−1}(G)} |S_h|.        (2)\n\nBy the induction hypothesis each summand |S_h| is divisible by ℓ, therefore their sum (2) is divisible by ℓ.  This completes the induction and proves part A. ∎\n\n\n\n————————————————————————\nPart B.  Sharpness of the exponent ν_ℓ(|C_r(G)|) = 1  \n\nWe produce, for every prime ℓ and every r ≥ 1, a finite group G with exactly one factor ℓ in |C_r(G)|.\n\n(i)  The trivial case r = 1.  \nTake G = C_ℓ.  Then C₁(G)=G and |C₁(G)|=ℓ, so ν_ℓ(|C₁(G)|)=1.\n\n(ii)  A uniform family for every r ≥ 2.  \nIf ℓ = 2 we take the symmetric group S₃; for odd ℓ we take the simple group  \n\n G = PSL₂(ℓ)   (of order ℓ(ℓ²−1)/2).\n\nStandard facts we use (see, e.g., Huppert–Blackburn or any text on PSL₂(q)):\n\n• Every non-identity ℓ-element of PSL₂(ℓ) lies in a unique cyclic subgroup of order ℓ, and its centraliser is precisely that subgroup.  \n• The number c of cyclic subgroups of order ℓ in PSL₂(ℓ) equals ℓ+1 (the number of points of the projective line over 𝔽_ℓ).\n\nExactly the same two facts hold for ℓ = 2 inside S₃ (c = 3).\n\nCounting commuting r-tuples.  \nTwo non-trivial ℓ-elements commute iff they belong to the same cyclic subgroup of order ℓ.  Consequently every member of C_r(G) is either the all-identity tuple E = (e,…,e) or else lies entirely inside a unique cyclic subgroup of order ℓ.  Inside one such subgroup H ≅ C_ℓ there are ℓ^r ordered r-tuples of ℓ-elements, of which precisely one is E.  Summing over the c subgroups we obtain  \n\n |C_r(G)| = 1 + c(ℓ^r − 1).             (3)\n\nSubstituting c = ℓ+1 gives  \n\n |C_r(G)| = 1 + (ℓ+1)(ℓ^r − 1)  \n     = (ℓ+1)ℓ^r − ℓ  \n     = ℓ · [(ℓ+1)ℓ^{\\,r−1} − 1].     (4)\n\nBecause the bracket is congruent to −1 modulo ℓ, it is not divisible by ℓ, so ν_ℓ(|C_r(G)|)=1.  This proves the sharpness for every r ≥ 2.\n\nCombining (i) and (ii) we have produced, for every prime ℓ and every r ≥ 1, a finite group G with ν_ℓ(|C_r(G)|)=1, completing part B. ∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.805489",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher-dimensional setting.  Instead of counting single ℓ-elements (r=1), the problem asks for the number of ordered r-tuples of pairwise-commuting ℓ-elements, dramatically enlarging the search space and the combinatorial complexity.\n\n2. Additional group-theoretic structure.  The proof must control centralisers of arbitrary tuples and repeatedly invoke Cauchy’s theorem inside those centralisers; simple orbit-counting no longer suffices.\n\n3. Multi-layer induction.  Each new coordinate introduces a fresh layer of centraliser analysis and a fresh application of Cauchy, so the argument grows in depth with r.  The original problem needs one counting trick; the enhanced variant needs r nested applications.\n\n4. Interacting concepts.  The solution blends three ideas: (i) centraliser containment, (ii) recursive extension of homomorphisms, and (iii) divisibility inherited from subgroups.  Any shortcut or naive counting overlooks the dependence of later coordinates on all previous ones.\n\n5. Stronger conclusion.  While the original theorem supplies a single factor of ℓ, the enhanced result forces ℓ^r—a power that increases with the dimension—making it strictly stronger for every r > 1 and unattainable by the original arguments alone."
      }
    },
    "original_kernel_variant": {
      "question": "Let ℓ be a prime and r ≥ 1 an integer.  \nFor a finite group G set  \n\n C_r(G)= { (g₁,…,g_r)∈G^r  |  g_i^ℓ = e for every i and g_i g_j = g_j g_i for all i,j }.  \n\nThus C_r(G) consists of the ordered r–tuples of pairwise–commuting ℓ-elements of G.\n\nA)  Assume that ℓ divides |G|.  Prove that |C_r(G)| is divisible by ℓ.  \n   (If ℓ∤|G| then C_r(G)= {(e,…,e)}, so no divisibility can be expected.)\n\nB)  Show that the exponent obtained in A) is best possible:  \n\n for every prime ℓ and every r ≥ 1 there exists a finite group G with  \n  ν_ℓ(|C_r(G)|)=1.\n\nA single, uniform proof is expected; avoid a long case–by–case discussion.  \n(Hints for B:  for ℓ=2 one may take the symmetric group S₃; for odd ℓ one may take the group PSL₂(ℓ).)",
      "solution": "Throughout let ℓ be a fixed prime, r ≥ 1 an integer, and G a finite group.\n\n————————————————————————\nPart A.  Divisibility of |C_r(G)| by ℓ (under ℓ | |G|)  \nWe argue by induction on r.\n\nBase step r = 1.  \nPut  \n\n T := {(x₀,…,x_{ℓ-1}) ∈ G^{ℓ}  |  x₀x₁···x_{ℓ-1}=e}.  \n\nClearly |T| = |G|^{ℓ-1}, which is divisible by ℓ because ℓ | |G|.  \nLet the cyclic group C_ℓ = ⟨σ⟩ act on T by the left shift  \n σ·(x₀,…,x_{ℓ-1}) = (x₁,…,x_{ℓ-1},x₀).  \nAll orbits have size ℓ except the fixed points, which satisfy x₀ = ··· = x_{ℓ-1} and hence are precisely the diagonal tuples (x,…,x) with x^ℓ = e.  The class equation gives  \n\n |T| ≡ # {x∈G | x^ℓ=e}  (mod ℓ).  \n\nBecause |T| is a multiple of ℓ, the set S₁(G):= {x∈G | x^ℓ = e} has cardinality divisible by ℓ.  Now  \n\n |C₁(G)| = |S₁(G)| = ℓ·k     for some k∈ℕ,\n\nso |C₁(G)| is a multiple of ℓ, establishing the base case.\n\nInduction step.  Assume r ≥ 2 and that the statement holds for r−1.  \nFor h=(h₁,…,h_{r−1})∈C_{r−1}(G) put H:=⟨h₁,…,h_{r−1}⟩; this is an elementary abelian ℓ-subgroup, hence H ≤ C_G(H).  Define  \n\n S_h := { g∈C_G(H) | g^ℓ=e }.        (1)\n\nApply the already–proved base case r = 1 to the finite group C_G(H): because ℓ divides |C_G(H)| (H is non-trivial unless h=(e,…,e), but the argument below also works for the trivial H), the set S_h has size divisible by ℓ.  Next observe that an r-tuple in C_r(G) is obtained by adjoining to h any element of S_h as its last coordinate, and every such choice is admissible.  Hence the map  \n\n C_{r−1}(G) × S_h  →  C_r(G),    (h,g) ↦ (h₁,…,h_{r−1},g)\n\nis bijective inside each fibre over h, yielding  \n\n |C_r(G)| = ∑_{h∈C_{r−1}(G)} |S_h|.        (2)\n\nBy the induction hypothesis each summand |S_h| is divisible by ℓ, therefore their sum (2) is divisible by ℓ.  This completes the induction and proves part A. ∎\n\n\n\n————————————————————————\nPart B.  Sharpness of the exponent ν_ℓ(|C_r(G)|) = 1  \n\nWe produce, for every prime ℓ and every r ≥ 1, a finite group G with exactly one factor ℓ in |C_r(G)|.\n\n(i)  The trivial case r = 1.  \nTake G = C_ℓ.  Then C₁(G)=G and |C₁(G)|=ℓ, so ν_ℓ(|C₁(G)|)=1.\n\n(ii)  A uniform family for every r ≥ 2.  \nIf ℓ = 2 we take the symmetric group S₃; for odd ℓ we take the simple group  \n\n G = PSL₂(ℓ)   (of order ℓ(ℓ²−1)/2).\n\nStandard facts we use (see, e.g., Huppert–Blackburn or any text on PSL₂(q)):\n\n• Every non-identity ℓ-element of PSL₂(ℓ) lies in a unique cyclic subgroup of order ℓ, and its centraliser is precisely that subgroup.  \n• The number c of cyclic subgroups of order ℓ in PSL₂(ℓ) equals ℓ+1 (the number of points of the projective line over 𝔽_ℓ).\n\nExactly the same two facts hold for ℓ = 2 inside S₃ (c = 3).\n\nCounting commuting r-tuples.  \nTwo non-trivial ℓ-elements commute iff they belong to the same cyclic subgroup of order ℓ.  Consequently every member of C_r(G) is either the all-identity tuple E = (e,…,e) or else lies entirely inside a unique cyclic subgroup of order ℓ.  Inside one such subgroup H ≅ C_ℓ there are ℓ^r ordered r-tuples of ℓ-elements, of which precisely one is E.  Summing over the c subgroups we obtain  \n\n |C_r(G)| = 1 + c(ℓ^r − 1).             (3)\n\nSubstituting c = ℓ+1 gives  \n\n |C_r(G)| = 1 + (ℓ+1)(ℓ^r − 1)  \n     = (ℓ+1)ℓ^r − ℓ  \n     = ℓ · [(ℓ+1)ℓ^{\\,r−1} − 1].     (4)\n\nBecause the bracket is congruent to −1 modulo ℓ, it is not divisible by ℓ, so ν_ℓ(|C_r(G)|)=1.  This proves the sharpness for every r ≥ 2.\n\nCombining (i) and (ii) we have produced, for every prime ℓ and every r ≥ 1, a finite group G with ν_ℓ(|C_r(G)|)=1, completing part B. ∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.615765",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher-dimensional setting.  Instead of counting single ℓ-elements (r=1), the problem asks for the number of ordered r-tuples of pairwise-commuting ℓ-elements, dramatically enlarging the search space and the combinatorial complexity.\n\n2. Additional group-theoretic structure.  The proof must control centralisers of arbitrary tuples and repeatedly invoke Cauchy’s theorem inside those centralisers; simple orbit-counting no longer suffices.\n\n3. Multi-layer induction.  Each new coordinate introduces a fresh layer of centraliser analysis and a fresh application of Cauchy, so the argument grows in depth with r.  The original problem needs one counting trick; the enhanced variant needs r nested applications.\n\n4. Interacting concepts.  The solution blends three ideas: (i) centraliser containment, (ii) recursive extension of homomorphisms, and (iii) divisibility inherited from subgroups.  Any shortcut or naive counting overlooks the dependence of later coordinates on all previous ones.\n\n5. Stronger conclusion.  While the original theorem supplies a single factor of ℓ, the enhanced result forces ℓ^r—a power that increases with the dimension—making it strictly stronger for every r > 1 and unattainable by the original arguments alone."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}