{
  "index": "1989-B-2",
  "type": "ALG",
  "tag": [
    "ALG"
  ],
  "difficulty": "",
  "question": "Let $S$ be a non-empty set with an associative operation that is left and\nright cancellative ($xy=xz$ implies $y=z$, and $yx=zx$ implies $y=z$).\nAssume that for every $a$ in $S$ the set $\\{a^n:\\,n=1, 2, 3, \\ldots\\}$ is\nfinite. Must $S$ be a group?",
  "solution": "Solution. Choose \\( a \\in S \\). The finiteness hypothesis implies that some term in the sequence \\( a, a^{2}, a^{3}, \\ldots \\) is repeated infinitely often, so we have \\( a^{m}=a^{n} \\) for some integers \\( m, n \\geq 1 \\) with \\( m-n \\geq 2 \\). Let \\( e=a^{m-n} \\). For any \\( x \\in S, a^{n} e x=a^{m} x \\), and cancelling \\( a^{n}=a^{m} \\) shows that \\( e x=x \\). Similarly \\( x e=x \\), so \\( e \\) is an identity. Now \\( a a^{m-n-1}=a^{m-n}=e \\) and \\( a^{m-n-1} a=a^{m-n}=e \\), so \\( a^{m-n-1} \\) is an inverse of \\( a \\). Since \\( S \\) is associative and has an identity, and since any \\( a \\in S \\) has an inverse, \\( S \\) is a group.",
  "vars": [
    "S",
    "a",
    "e",
    "x",
    "y",
    "z"
  ],
  "params": [
    "m",
    "n"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "S": "setspace",
        "a": "basiselem",
        "x": "arbitrx",
        "y": "arbitry",
        "z": "arbitrz",
        "m": "exponentm",
        "n": "exponentn"
      },
      "question": "Let $setspace$ be a non-empty set with an associative operation that is left and right cancellative ($arbitrx\\,\\arbitry=\\arbitrx\\,\\arbitrz$ implies $\\arbitry=\\arbitrz$, and $\\arbitry\\,\\arbitrx=\\arbitrz\\,\\arbitrx$ implies $\\arbitry=\\arbitrz$). Assume that for every $basiselem$ in $setspace$ the set $\\{basiselem^{exponentn}:\\,exponentn=1, 2, 3, \\ldots\\}$ is finite. Must $setspace$ be a group?",
      "solution": "Solution. Choose $basiselem \\in setspace$. The finiteness hypothesis implies that some term in the sequence $basiselem, basiselem^{2}, basiselem^{3}, \\ldots$ is repeated infinitely often, so we have $basiselem^{exponentm}=basiselem^{exponentn}$ for some integers $exponentm, exponentn \\ge 1$ with $exponentm-exponentn \\ge 2$. Let $e=basiselem^{exponentm-exponentn}$. For any $arbitrx \\in setspace$, $basiselem^{exponentn} e \\; arbitrx=basiselem^{exponentm} \\; arbitrx$, and cancelling $basiselem^{exponentn}=basiselem^{exponentm}$ shows that $e \\; arbitrx=arbitrx$. Similarly $arbitrx \\; e=arbitrx$, so $e$ is an identity. Now $basiselem \\; basiselem^{exponentm-exponentn-1}=basiselem^{exponentm-exponentn}=e$ and $basiselem^{exponentm-exponentn-1} \\; basiselem=basiselem^{exponentm-exponentn}=e$, so $basiselem^{exponentm-exponentn-1}$ is an inverse of $basiselem$. Since $setspace$ is associative and has an identity, and since any $basiselem \\in setspace$ has an inverse, $setspace$ is a group."
    },
    "descriptive_long_confusing": {
      "map": {
        "S": "magazine",
        "a": "lamplight",
        "e": "vineyard",
        "x": "handrail",
        "y": "cupboard",
        "z": "platform",
        "m": "bluebird",
        "n": "necklace"
      },
      "question": "Let $magazine$ be a non-empty set with an associative operation that is left and\nright cancellative ($handrail cupboard = handrail platform$ implies $cupboard = platform$, and $cupboard handrail = platform handrail$ implies $cupboard = platform$).\nAssume that for every $lamplight$ in $magazine$ the set $\\{lamplight^{necklace}:\\,necklace=1, 2, 3, \\ldots\\}$ is\nfinite. Must $magazine$ be a group?",
      "solution": "Solution. Choose \\( lamplight \\in magazine \\). The finiteness hypothesis implies that some term in the sequence \\( lamplight, lamplight^{2}, lamplight^{3}, \\ldots \\) is repeated infinitely often, so we have \\( lamplight^{bluebird}=lamplight^{necklace} \\) for some integers \\( bluebird, necklace \\geq 1 \\) with \\( bluebird-necklace \\geq 2 \\). Let \\( vineyard=lamplight^{bluebird-necklace} \\). For any \\( handrail \\in magazine, lamplight^{necklace} vineyard handrail=lamplight^{bluebird} handrail \\), and cancelling \\( lamplight^{necklace}=lamplight^{bluebird} \\) shows that \\( vineyard handrail = handrail \\). Similarly \\( handrail vineyard = handrail \\), so \\( vineyard \\) is an identity. Now \\( lamplight lamplight^{bluebird-necklace-1}=lamplight^{bluebird-necklace}=vineyard \\) and \\( lamplight^{bluebird-necklace-1} lamplight=lamplight^{bluebird-necklace}=vineyard \\), so \\( lamplight^{bluebird-necklace-1} \\) is an inverse of \\( lamplight \\). Since \\( magazine \\) is associative and has an identity, and since any \\( lamplight \\in magazine \\) has an inverse, \\( magazine \\) is a group."
    },
    "descriptive_long_misleading": {
      "map": {
        "S": "emptiness",
        "a": "totality",
        "e": "alterity",
        "x": "constant",
        "y": "knownvalue",
        "z": "fixedvalue",
        "m": "randomness",
        "n": "chaosness"
      },
      "question": "Let $emptiness$ be a non-empty set with an associative operation that is left and\nright cancellative ($constant knownvalue = constant fixedvalue$ implies $knownvalue=fixedvalue$, and $knownvalue constant = fixedvalue constant$ implies $knownvalue=fixedvalue$).\nAssume that for every $totality$ in $emptiness$ the set $\\{totality^{chaosness}:\\,\\chaosness=1, 2, 3, \\ldots\\}$ is\nfinite. Must $emptiness$ be a group?",
      "solution": "Solution. Choose \\( totality \\in emptiness \\). The finiteness hypothesis implies that some term in the sequence \\( totality, totality^{2}, totality^{3}, \\ldots \\) is repeated infinitely often, so we have \\( totality^{randomness}=totality^{chaosness} \\) for some integers \\( randomness, chaosness \\geq 1 \\) with \\( randomness-chaosness \\geq 2 \\). Let \\( alterity=totality^{randomness-chaosness} \\). For any \\( constant \\in emptiness, totality^{chaosness} alterity constant=totality^{randomness} constant \\), and cancelling \\( totality^{chaosness}=totality^{randomness} \\) shows that \\( alterity constant=constant \\). Similarly \\( constant alterity=constant \\), so \\( alterity \\) is an identity. Now \\( totality totality^{randomness-chaosness-1}=totality^{randomness-chaosness}=alterity \\) and \\( totality^{randomness-chaosness-1} totality=totality^{randomness-chaosness}=alterity \\), so \\( totality^{randomness-chaosness-1} \\) is an inverse of \\( totality \\). Since \\( emptiness \\) is associative and has an identity, and since any \\( totality \\in emptiness \\) has an inverse, \\( emptiness \\) is a group."
    },
    "garbled_string": {
      "map": {
        "S": "qzxwvtnp",
        "a": "hjgrksla",
        "e": "plsdmcnv",
        "x": "vrtklmwa",
        "y": "jkdphzsu",
        "z": "gqvnrcye",
        "m": "ufdsbqei",
        "n": "rkeotmza"
      },
      "question": "Let $qzxwvtnp$ be a non-empty set with an associative operation that is left and\nright cancellative ($vrtklmwa jkdphzsu=vrtklmwa gqvnrcye$ implies $jkdphzsu=gqvnrcye$, and $jkdphzsu vrtklmwa=gqvnrcye vrtklmwa$ implies $jkdphzsu=gqvnrcye$).\nAssume that for every $hjgrksla$ in $qzxwvtnp$ the set $\\{hjgrksla^{rkeotmza}:\\,rkeotmza=1, 2, 3, \\ldots\\}$ is\nfinite. Must $qzxwvtnp$ be a group?",
      "solution": "Solution. Choose \\( hjgrksla \\in qzxwvtnp \\). The finiteness hypothesis implies that some term in the sequence \\( hjgrksla, hjgrksla^{2}, hjgrksla^{3}, \\ldots \\) is repeated infinitely often, so we have \\( hjgrksla^{ufdsbqei}=hjgrksla^{rkeotmza} \\) for some integers \\( ufdsbqei, rkeotmza \\geq 1 \\) with \\( ufdsbqei-rkeotmza \\geq 2 \\). Let \\( plsdmcnv=hjgrksla^{ufdsbqei-rkeotmza} \\). For any \\( vrtklmwa \\in qzxwvtnp, hjgrksla^{rkeotmza} plsdmcnv vrtklmwa = hjgrksla^{ufdsbqei} vrtklmwa \\), and cancelling \\( hjgrksla^{rkeotmza}=hjgrksla^{ufdsbqei} \\) shows that \\( plsdmcnv vrtklmwa=vrtklmwa \\). Similarly \\( vrtklmwa plsdmcnv=vrtklmwa \\), so \\( plsdmcnv \\) is an identity. Now \\( hjgrksla hjgrksla^{ufdsbqei-rkeotmza-1}=hjgrksla^{ufdsbqei-rkeotmza}=plsdmcnv \\) and \\( hjgrksla^{ufdsbqei-rkeotmza-1} hjgrksla=hjgrksla^{ufdsbqei-rkeotmza}=plsdmcnv \\), so \\( hjgrksla^{ufdsbqei-rkeotmza-1} \\) is an inverse of \\( hjgrksla \\). Since \\( qzxwvtnp \\) is associative and has an identity, and since any \\( hjgrksla \\in qzxwvtnp \\) has an inverse, \\( qzxwvtnp \\) is a group."
    },
    "kernel_variant": {
      "question": "Let \\(S\\) be a non-empty cancellative semigroup; that is, \\(\\cdot\\) is associative on \\(S\\) and whenever \\(x\\cdot y = x\\cdot z\\) or \\(y\\cdot x = z\\cdot x\\) we have \\(y=z\\).   \nFor every element \\(a\\in S\\) assume that the sequence of positive powers of \\(a\\) is not injective: there exist integers\n\\[1\\le r<s, \\; s-r\\ge 3,\\qquad\\text{with }\\qquad a^{\\,r}=a^{\\,s}.\\]\nProve that \\(S\\) is a group (i.e. possesses an identity element \\(1_{S}\\) and every element has a two-sided inverse).",
      "solution": "Fix an arbitrary element \\(a\\in S\\).  By hypothesis choose integers \\(r,s\\) with \\(1\\le r<s\\) and \\(s-r\\ge 3\\) such that\n\\[a^{r}=a^{s}.\\]\nSet\n\\[k:=s-r\\;(\\ge 3), \\qquad 1_{S}:=a^{k}.\\]\n------------------------------------------------------\n1.  \\(1_{S}\\) is a two-sided identity.\nFor any \\(x\\in S\\), associativity gives\n\\[a^{r}\\,1_{S}\\,x= a^{r}a^{k}x=a^{s}x.\\]\nBecause \\(a^{r}=a^{s}\\), we also have \\(a^{r}x=a^{s}x\\), so\n\\[a^{r}(1_{S}x)=a^{r}x.\\]\nLeft cancellation then yields\n\\[1_{S}\\,x=x.\\]\nA symmetric argument with \\(x\\,1_{S}\\) and right cancellation shows \\(x\\,1_{S}=x\\).  Thus \\(1_{S}\\) acts as an identity on all of \\(S\\).\n------------------------------------------------------\n2.  Every element has an inverse.\nFor our fixed \\(a\\) put\n\\[a^{\\#}:=a^{k-1}\\;(\\text{note }k-1\\ge 2).\\]\nThen\n\\[a\\,a^{\\#}=a\\,a^{k-1}=a^{k}=1_{S}\n\\quad\\text{and}\\quad\na^{\\#}a=a^{k-1}a=a^{k}=1_{S},\\]\nso \\(a^{\\#}\\) is a two-sided inverse of \\(a\\).\nBecause \\(a\\in S\\) was arbitrary, the same construction (using the appropriate \\(r,s\\) for each element) furnishes a two-sided inverse for every member of \\(S\\).\n------------------------------------------------------\n3.  Conclusion.\nThe semigroup \\(S\\) is associative by assumption, possesses the identity \\(1_{S}\\) found in step 1, and every element has an inverse by step 2.  Therefore \\(S\\) is a group.\n\n\\boxed{}",
      "_meta": {
        "core_steps": [
          "Finiteness of the power–orbit ⇒ ∃ m>n with a^m = a^n (pigeon-hole).",
          "Let e := a^{m−n}; left/right cancellation gives e·x = x and x·e = x for every x.",
          "Take a^{m−n−1} as a^{-1}; verify a·a^{-1}=a^{-1}·a=e via associativity.",
          "Identity + inverses for all elements ⇒ S is a group."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "The exact finiteness statement can be weakened to any hypothesis guaranteeing a repetition of powers (e.g. ‘the sequence a, a², … is not injective’).",
            "original": "for every a in S, {a^n : n ≥ 1} is finite"
          },
          "slot2": {
            "description": "Only a gap of at least 1 is needed; the proof picks m−n ≥ 2 merely to keep all exponents positive.",
            "original": "choose m, n with m−n ≥ 2"
          },
          "slot3": {
            "description": "Notation for the prospective identity and inverse can be changed (e.g. call them 1_S and a′ ).",
            "original": "e = a^{m−n}; inverse = a^{m−n−1}"
          },
          "slot4": {
            "description": "The two separate cancellation clauses can be consolidated into the single phrase ‘S is cancellative’.",
            "original": "left and right cancellative stated separately"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}