{
  "index": "1977-B-6",
  "type": "ALG",
  "tag": [
    "ALG",
    "NT"
  ],
  "difficulty": "",
  "question": "Problem B-6\nLet \\( H \\) be a subgroup with \\( h \\) elements in a group \\( G \\). Suppose that \\( G \\) has an element \\( a \\) such that for all \\( x \\) in \\( H \\), \\( (x a)^{3}=1 \\), the identity. In \\( G \\), let \\( P \\) be the subset of all products \\( x_{1} a x_{2} a \\cdots x_{n} a \\), with \\( n \\) a positive integer and the \\( x_{i} \\) in H.\n(a) Show that \\( P \\) is a finite set.\n(b) Show that, in fact, \\( P \\) has no more than \\( 3 h^{2} \\) elements.",
  "solution": "B-6.\nClearly \\( 1 \\in H \\). Also \\( x \\in H \\) implies \\( x^{-1} \\in H \\). Then the hypothesis implies that \\( a^{-1}=a^{2} \\) and that xaxaxa \\( =1=x^{-1} a x^{-1} a x^{-1} a \\) when \\( x \\in H \\). Thus one easily shows that\n(i) \\( a x a=x^{-1} a^{2} x^{-1} \\),\n(ii) \\( a^{2} x a^{2}=x^{-1} a x^{-1} \\).\n\nLet\n\\[\n\\begin{array}{ll}\nA=\\{x a y: x, y \\in H\\}, & B=\\left\\{x a^{2} y: x, y \\in H\\right\\}, \\\\\nC=\\left\\{x a^{2} y a: x, y \\in H\\right\\}, & \\text { and } \\quad Q=A \\cup B \\cup C .\n\\end{array}\n\\]\n\nEach of \\( A, B, C \\) has at most \\( h^{2} \\) elements; hence \\( Q \\) has at most \\( 3 h^{2} \\) elements. Thus it suffices to prove that \\( x_{1} a x_{2} a \\cdots x_{n} a \\in Q \\) when each \\( x_{i} \\in H \\). We do this by induction on \\( n \\).\n\nFor \\( n=1 \\), one sees that \\( x_{1} a=x_{1} a \\cdot 1 \\in A \\subseteq Q \\). Now let \\( x_{1}, x_{2}, \\ldots, x_{k+1} \\in H ; x_{1} a x_{2} a \\cdots x_{k} a= \\) \\( q, x_{k+1}=z \\), and \\( q z a=p \\). Inductively, we assume \\( q \\in Q \\) and seek to show \\( p \\in Q \\). The assumption implies that \\( q \\) is in \\( A, B \\), or \\( C \\). If \\( q=x a y \\in A \\), then\n\\[\np=(x a y) z a=x a(y z) a=x(y z)^{-1} a^{2}(y z)^{-1} \\in B \\subseteq Q,\n\\]\nusing (i). If \\( q=x a^{2} y \\in B \\), then \\( p=x a^{2} y z a \\in C \\subseteq Q \\). If \\( q=x a^{2} y a \\in C \\), then\n\\[\n\\begin{aligned}\np & =x a^{2} y(a z a)=x a^{2} y\\left(z^{-1} a^{2} z^{-1}\\right)=x\\left[a^{2}\\left(y z^{-1}\\right) a^{2}\\right] z^{-1} \\\\\n& =x\\left(y z^{-1}\\right)^{-1} a\\left(y z^{-1}\\right)^{-1} z^{-1} \\in A \\subseteq Q,\n\\end{aligned}\n\\]\nusing (i) and (ii).",
  "vars": [
    "x",
    "n",
    "x_i",
    "x_1",
    "x_2",
    "x_n",
    "y",
    "z",
    "q",
    "p",
    "k",
    "x_k",
    "x_k+1"
  ],
  "params": [
    "G",
    "H",
    "h",
    "a",
    "P",
    "A",
    "B",
    "C",
    "Q"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "elemvar",
        "n": "indexn",
        "x_i": "elemindi",
        "x_1": "elemone",
        "x_2": "elemtwo",
        "x_n": "elemlast",
        "y": "secondy",
        "z": "thirdz",
        "q": "quasiq",
        "p": "plimit",
        "k": "indexk",
        "x_k": "elemk",
        "x_k+1": "elemnxt",
        "G": "grandg",
        "H": "subgrp",
        "h": "subcnt",
        "a": "specele",
        "P": "prodset",
        "A": "setalpha",
        "B": "setbeta",
        "C": "setgamma",
        "Q": "settheta"
      },
      "question": "Problem B-6\nLet \\( subgrp \\) be a subgroup with \\( subcnt \\) elements in a group \\( grandg \\). Suppose that \\( grandg \\) has an element \\( specele \\) such that for all \\( elemvar \\) in \\( subgrp \\), \\( (elemvar\\, specele)^{3}=1 \\), the identity. In \\( grandg \\), let \\( prodset \\) be the subset of all products \\( elemone\\, specele\\, elemtwo\\, specele \\cdots elemlast\\, specele \\), with \\( indexn \\) a positive integer and the \\( elemindi \\) in \\( subgrp \\).\n(a) Show that \\( prodset \\) is a finite set.\n(b) Show that, in fact, \\( prodset \\) has no more than \\( 3 subcnt^{2} \\) elements.",
      "solution": "B-6.\nClearly \\( 1 \\in subgrp \\). Also \\( elemvar \\in subgrp \\) implies \\( elemvar^{-1} \\in subgrp \\). Then the hypothesis implies that \\( specele^{-1}=specele^{2} \\) and that elemvarspeceleelemvarspeceleelemvarspecele \\( =1=elemvar^{-1}\\, specele\\, elemvar^{-1}\\, specele\\, elemvar^{-1}\\, specele \\) when \\( elemvar \\in subgrp \\). Thus one easily shows that\n(i) \\( specele\\, elemvar\\, specele=elemvar^{-1}\\, specele^{2}\\, elemvar^{-1} \\),\n(ii) \\( specele^{2}\\, elemvar\\, specele^{2}=elemvar^{-1}\\, specele\\, elemvar^{-1} \\).\n\nLet\n\\[\n\\begin{array}{ll}\nsetalpha=\\{elemvar\\, specele\\, secondy: elemvar, secondy \\in subgrp\\}, & setbeta=\\left\\{elemvar\\, specele^{2}\\, secondy: elemvar, secondy \\in subgrp\\right\\}, \\\\\nsetgamma=\\left\\{elemvar\\, specele^{2}\\, secondy\\, specele: elemvar, secondy \\in subgrp\\right\\}, & \\text { and } \\quad settheta=setalpha \\cup setbeta \\cup setgamma .\n\\end{array}\n\\]\n\nEach of \\( setalpha, setbeta, setgamma \\) has at most \\( subcnt^{2} \\) elements; hence \\( settheta \\) has at most \\( 3 subcnt^{2} \\) elements. Thus it suffices to prove that \\( elemone\\, specele\\, elemtwo\\, specele \\cdots elemlast\\, specele \\in settheta \\) when each \\( elemindi \\in subgrp \\). We do this by induction on \\( indexn \\).\n\nFor \\( indexn=1 \\), one sees that \\( elemone\\, specele=elemone\\, specele \\cdot 1 \\in setalpha \\subseteq settheta \\). Now let \\( elemone, elemtwo, \\ldots, elemnxt \\in subgrp ; elemone\\, specele\\, elemtwo\\, specele \\cdots elemk\\, specele= \\) \\( quasiq, elemnxt=thirdz \\), and \\( quasiq\\, thirdz\\, specele=plimit \\). Inductively, we assume \\( quasiq \\in settheta \\) and seek to show \\( plimit \\in settheta \\). The assumption implies that \\( quasiq \\) is in \\( setalpha, setbeta \\), or \\( setgamma \\). If \\( quasiq=elemvar\\, specele\\, secondy \\in setalpha \\), then\n\\[\nplimit=(elemvar\\, specele\\, secondy)\\, thirdz\\, specele=elemvar\\, specele(secondy\\, thirdz)\\, specele=elemvar(secondy\\, thirdz)^{-1}\\, specele^{2}(secondy\\, thirdz)^{-1} \\in setbeta \\subseteq settheta,\n\\]\nusing (i). If \\( quasiq=elemvar\\, specele^{2}\\, secondy \\in setbeta \\), then \\( plimit=elemvar\\, specele^{2}\\, secondy\\, thirdz\\, specele \\in setgamma \\subseteq settheta \\). If \\( quasiq=elemvar\\, specele^{2}\\, secondy\\, specele \\in setgamma \\), then\n\\[\n\\begin{aligned}\nplimit & =elemvar\\, specele^{2}\\, secondy(specele\\, thirdz\\, specele)=elemvar\\, specele^{2}\\, secondy\\left(thirdz^{-1}\\, specele^{2}\\, thirdz^{-1}\\right)=elemvar\\left[specele^{2}\\left(secondy\\, thirdz^{-1}\\right)\\, specele^{2}\\right]\\, thirdz^{-1} \\\\\n& =elemvar\\left(secondy\\, thirdz^{-1}\\right)^{-1}\\, specele\\left(secondy\\, thirdz^{-1}\\right)^{-1}\\, thirdz^{-1} \\in setalpha \\subseteq settheta,\n\\end{aligned}\n\\]\nusing (i) and (ii)."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "peppermint",
        "n": "lemonade",
        "x_i": "tangerine",
        "x_1": "crocodile",
        "x_2": "chandelier",
        "x_n": "silverspoon",
        "y": "bookshelf",
        "z": "drumstick",
        "q": "cloudburst",
        "p": "rainshower",
        "k": "marshmallow",
        "x_k": "gingerbread",
        "x_k+1": "butternut",
        "G": "hummingbird",
        "H": "watermelon",
        "h": "wisteria",
        "a": "gravestone",
        "P": "sandstone",
        "A": "blueberry",
        "B": "peanutbutter",
        "C": "masterwork",
        "Q": "dragonfly"
      },
      "question": "Problem B-6\nLet \\( watermelon \\) be a subgroup with \\( wisteria \\) elements in a group \\( hummingbird \\). Suppose that \\( hummingbird \\) has an element \\( gravestone \\) such that for all \\( peppermint \\) in \\( watermelon \\), \\( (peppermint gravestone)^{3}=1 \\), the identity. In \\( hummingbird \\), let \\( sandstone \\) be the subset of all products \\( crocodile gravestone chandelier gravestone \\cdots silverspoon gravestone \\), with \\( lemonade \\) a positive integer and the \\( tangerine \\) in watermelon.\n(a) Show that \\( sandstone \\) is a finite set.\n(b) Show that, in fact, \\( sandstone \\) has no more than \\( 3 wisteria^{2} \\) elements.",
      "solution": "B-6.\nClearly \\( 1 \\in watermelon \\). Also \\( peppermint \\in watermelon \\) implies \\( peppermint^{-1} \\in watermelon \\). Then the hypothesis implies that \\( gravestone^{-1}=gravestone^{2} \\) and that peppermintgravestonepeppermintgravestonepeppermintgravestone \\( =1=peppermint^{-1} gravestone peppermint^{-1} gravestone peppermint^{-1} gravestone \\) when \\( peppermint \\in watermelon \\). Thus one easily shows that\n(i) \\( gravestone peppermint gravestone=peppermint^{-1} gravestone^{2} peppermint^{-1} \\),\n(ii) \\( gravestone^{2} peppermint gravestone^{2}=peppermint^{-1} gravestone peppermint^{-1} \\).\n\nLet\n\\[\n\\begin{array}{ll}\nblueberry=\\{peppermint gravestone bookshelf: peppermint, bookshelf \\in watermelon\\}, & peanutbutter=\\left\\{peppermint gravestone^{2} bookshelf: peppermint, bookshelf \\in watermelon\\right\\}, \\\\\nmasterwork=\\left\\{peppermint gravestone^{2} bookshelf gravestone: peppermint, bookshelf \\in watermelon\\right\\}, & \\text { and } \\quad dragonfly=blueberry \\cup peanutbutter \\cup masterwork .\n\\end{array}\n\\]\n\nEach of \\( blueberry, peanutbutter, masterwork \\) has at most \\( wisteria^{2} \\) elements; hence \\( dragonfly \\) has at most \\( 3 wisteria^{2} \\) elements. Thus it suffices to prove that \\( crocodile gravestone chandelier gravestone \\cdots silverspoon gravestone \\in dragonfly \\) when each \\( tangerine \\in watermelon \\). We do this by induction on \\( lemonade \\).\n\nFor \\( lemonade=1 \\), one sees that \\( crocodile gravestone=crocodile gravestone \\cdot 1 \\in blueberry \\subseteq dragonfly \\). Now let \\( crocodile, chandelier, \\ldots, gingerbread, butternut \\in watermelon ; crocodile gravestone chandelier gravestone \\cdots gingerbread gravestone= cloudburst, butternut=drumstick \\), and \\( cloudburst drumstick gravestone=rainshower \\). Inductively, we assume \\( cloudburst \\in dragonfly \\) and seek to show \\( rainshower \\in dragonfly \\). The assumption implies that \\( cloudburst \\) is in \\( blueberry, peanutbutter \\), or \\( masterwork \\). If \\( cloudburst=peppermint gravestone bookshelf \\in blueberry \\), then\n\\[\nrainshower=(peppermint gravestone bookshelf) drumstick gravestone=peppermint gravestone(bookshelf drumstick) gravestone=peppermint(bookshelf drumstick)^{-1} gravestone^{2}(bookshelf drumstick)^{-1} \\in peanutbutter \\subseteq dragonfly,\n\\]\nusing (i). If \\( cloudburst=peppermint gravestone^{2} bookshelf \\in peanutbutter \\), then \\( rainshower=peppermint gravestone^{2} bookshelf drumstick gravestone \\in masterwork \\subseteq dragonfly \\). If \\( cloudburst=peppermint gravestone^{2} bookshelf gravestone \\in masterwork \\), then\n\\[\n\\begin{aligned}\nrainshower & =peppermint gravestone^{2} bookshelf(gravestone drumstick gravestone)=peppermint gravestone^{2} bookshelf\\left(drumstick^{-1} gravestone^{2} drumstick^{-1}\\right)=peppermint\\left[gravestone^{2}\\left(bookshelf drumstick^{-1}\\right) gravestone^{2}\\right] drumstick^{-1} \\\\\n& =peppermint\\left(bookshelf drumstick^{-1}\\right)^{-1} gravestone\\left(bookshelf drumstick^{-1}\\right)^{-1} drumstick^{-1} \\in blueberry \\subseteq dragonfly,\n\\end{aligned}\n\\]\nusing (i) and (ii)."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "exteriorelement",
        "n": "trivialdigit",
        "x_i": "outsiderindex",
        "x_1": "lastmember",
        "x_2": "penultmember",
        "x_n": "firstmember",
        "y": "foreignvalue",
        "z": "strangerunit",
        "q": "divisorprod",
        "p": "factorresult",
        "k": "constantmark",
        "x_k": "fixedmember",
        "x_k+1": "nextmember",
        "G": "nongrouping",
        "H": "outsideset",
        "h": "infinitesize",
        "a": "identityelem",
        "P": "superset",
        "A": "complement",
        "B": "contraryset",
        "C": "exteriorset",
        "Q": "totality"
      },
      "question": "Problem B-6\nLet \\( outsideset \\) be a subgroup with \\( infinitesize \\) elements in a group \\( nongrouping \\). Suppose that \\( nongrouping \\) has an element \\( identityelem \\) such that for all \\( exteriorelement \\) in outsideset, \\( (exteriorelement\\ identityelem)^{3}=1 \\), the identity. In \\( nongrouping \\), let \\( superset \\) be the subset of all products \\( lastmember\\ identityelem\\ penultmember\\ identityelem \\cdots firstmember\\ identityelem \\), with \\( trivialdigit \\) a positive integer and the \\( outsiderindex \\) in outsideset.\n(a) Show that \\( superset \\) is a finite set.\n(b) Show that, in fact, \\( superset \\) has no more than \\( 3\\ infinitesize^{2} \\) elements.",
      "solution": "B-6.\nClearly \\( 1 \\in outsideset \\). Also \\( exteriorelement \\in outsideset \\) implies \\( exteriorelement^{-1} \\in outsideset \\). Then the hypothesis implies that \\( identityelem^{-1}=identityelem^{2} \\) and that exteriorelement identityelem exteriorelement identityelem exteriorelement identityelem \\( =1=exteriorelement^{-1} identityelem exteriorelement^{-1} identityelem exteriorelement^{-1} identityelem \\) when \\( exteriorelement \\in outsideset \\). Thus one easily shows that\n(i) \\( identityelem\\ exteriorelement\\ identityelem = exteriorelement^{-1}\\ identityelem^{2}\\ exteriorelement^{-1} \\),\n(ii) \\( identityelem^{2}\\ exteriorelement\\ identityelem^{2} = exteriorelement^{-1}\\ identityelem\\ exteriorelement^{-1} \\).\n\nLet\n\\[\n\\begin{array}{ll}\ncomplement=\\{exteriorelement\\ identityelem\\ foreignvalue: exteriorelement, foreignvalue \\in outsideset\\}, & contraryset=\\left\\{exteriorelement\\ identityelem^{2}\\ foreignvalue: exteriorelement, foreignvalue \\in outsideset\\right\\}, \\\\\nexteriorset=\\left\\{exteriorelement\\ identityelem^{2}\\ foreignvalue\\ identityelem: exteriorelement, foreignvalue \\in outsideset\\right\\}, & \\text { and } \\quad totality=complement \\cup contraryset \\cup exteriorset .\n\\end{array}\n\\]\n\nEach of \\( complement, contraryset, exteriorset \\) has at most \\( infinitesize^{2} \\) elements; hence \\( totality \\) has at most \\( 3\\ infinitesize^{2} \\) elements. Thus it suffices to prove that \\( lastmember\\ identityelem\\ penultmember\\ identityelem \\cdots firstmember\\ identityelem \\in totality \\) when each \\( outsiderindex \\in outsideset \\). We do this by induction on \\( trivialdigit \\).\n\nFor \\( trivialdigit=1 \\), one sees that \\( lastmember\\ identityelem = lastmember\\ identityelem \\cdot 1 \\in complement \\subseteq totality \\). Now let \\( lastmember, penultmember, \\ldots, nextmember \\in outsideset ; lastmember\\ identityelem\\ penultmember\\ identityelem \\cdots fixedmember\\ identityelem = divisorprod, fixedmember = strangerunit \\), and \\( divisorprod\\ strangerunit\\ identityelem = factorresult \\). Inductively, we assume \\( divisorprod \\in totality \\) and seek to show \\( factorresult \\in totality \\). The assumption implies that \\( divisorprod \\) is in \\( complement, contraryset \\), or \\( exteriorset \\). If \\( divisorprod = exteriorelement\\ identityelem\\ foreignvalue \\in complement \\), then\n\\[\nfactorresult = (exteriorelement\\ identityelem\\ foreignvalue)\\ strangerunit\\ identityelem = exteriorelement\\ identityelem (foreignvalue\\ strangerunit)\\ identityelem = exteriorelement (foreignvalue\\ strangerunit)^{-1}\\ identityelem^{2} (foreignvalue\\ strangerunit)^{-1} \\in contraryset \\subseteq totality,\n\\]\nusing (i). If \\( divisorprod = exteriorelement\\ identityelem^{2}\\ foreignvalue \\in contraryset \\), then \\( factorresult = exteriorelement\\ identityelem^{2}\\ foreignvalue\\ strangerunit\\ identityelem \\in exteriorset \\subseteq totality \\). If \\( divisorprod = exteriorelement\\ identityelem^{2}\\ foreignvalue\\ identityelem \\in exteriorset \\), then\n\\[\n\\begin{aligned}\nfactorresult & = exteriorelement\\ identityelem^{2}\\ foreignvalue (identityelem\\ strangerunit\\ identityelem)= exteriorelement\\ identityelem^{2}\\ foreignvalue \\left( strangerunit^{-1}\\ identityelem^{2}\\ strangerunit^{-1} \\right)= exteriorelement \\left[ identityelem^{2}\\left( foreignvalue\\ strangerunit^{-1} \\right) identityelem^{2} \\right] strangerunit^{-1} \\\\\n& = exteriorelement \\left( foreignvalue\\ strangerunit^{-1} \\right)^{-1}\\ identityelem \\left( foreignvalue\\ strangerunit^{-1} \\right)^{-1}\\ strangerunit^{-1} \\in complement \\subseteq totality,\n\\end{aligned}\n\\]\nusing (i) and (ii)."
    },
    "garbled_string": {
      "map": {
        "x": "mnbvcxzq",
        "n": "poiulkjh",
        "x_i": "qwertyui",
        "x_1": "zlkjhgfd",
        "x_2": "asdfghjk",
        "x_n": "qazxswed",
        "y": "edcrfvtg",
        "z": "rfvtgbyh",
        "q": "tgbznhyu",
        "p": "yhnujmik",
        "k": "ikmjnhbg",
        "x_k": "uytredcv",
        "x_k+1": "jnhtgbrf",
        "G": "wsxedcrf",
        "H": "vfrtgbnh",
        "h": "olpuytre",
        "a": "plokijuh",
        "P": "mkoijnuh",
        "A": "bnhgvrft",
        "B": "cftgvbnh",
        "C": "drtyhnbg",
        "Q": "esfregtb"
      },
      "question": "Problem B-6\nLet \\( vfrtgbnh \\) be a subgroup with \\( olpuytre \\) elements in a group \\( wsxedcrf \\). Suppose that \\( wsxedcrf \\) has an element \\( plokijuh \\) such that for all \\( mnbvcxzq \\) in \\( vfrtgbnh \\), \\( (mnbvcxzq plokijuh)^{3}=1 \\), the identity. In \\( wsxedcrf \\), let \\( mkoijnuh \\) be the subset of all products \\( zlkjhgfd plokijuh asdfghjk plokijuh \\cdots qazxswed plokijuh \\), with \\( poiulkjh \\) a positive integer and the \\( qwertyui \\) in \\( vfrtgbnh \\).\n(a) Show that \\( mkoijnuh \\) is a finite set.\n(b) Show that, in fact, \\( mkoijnuh \\) has no more than \\( 3 olpuytre^{2} \\) elements.",
      "solution": "B-6.\nClearly \\( 1 \\in vfrtgbnh \\). Also \\( mnbvcxzq \\in vfrtgbnh \\) implies \\( mnbvcxzq^{-1} \\in vfrtgbnh \\). Then the hypothesis implies that \\( plokijuh^{-1}=plokijuh^{2} \\) and that mnbvcxzqplokijuhmnbvcxzqplokijuhmnbvcxzqplokijuh \\( =1=mnbvcxzq^{-1} plokijuh mnbvcxzq^{-1} plokijuh mnbvcxzq^{-1} plokijuh \\) when \\( mnbvcxzq \\in vfrtgbnh \\). Thus one easily shows that\n(i) \\( plokijuh mnbvcxzq plokijuh=mnbvcxzq^{-1} plokijuh^{2} mnbvcxzq^{-1} \\),\n(ii) \\( plokijuh^{2} mnbvcxzq plokijuh^{2}=mnbvcxzq^{-1} plokijuh mnbvcxzq^{-1} \\).\n\nLet\n\\[\n\\begin{array}{ll}\nbnhgvrft=\\{mnbvcxzq plokijuh edcrfvtg: mnbvcxzq, edcrfvtg \\in vfrtgbnh\\}, & cftgvbnh=\\left\\{mnbvcxzq plokijuh^{2} edcrfvtg: mnbvcxzq, edcrfvtg \\in vfrtgbnh\\right\\}, \\\\\ndrtyhnbg=\\left\\{mnbvcxzq plokijuh^{2} edcrfvtg plokijuh: mnbvcxzq, edcrfvtg \\in vfrtgbnh\\right\\}, & \\text { and } \\quad esfregtb=bnhgvrft \\cup cftgvbnh \\cup drtyhnbg .\n\\end{array}\n\\]\n\nEach of \\( bnhgvrft, cftgvbnh, drtyhnbg \\) has at most \\( olpuytre^{2} \\) elements; hence \\( esfregtb \\) has at most \\( 3 olpuytre^{2} \\) elements. Thus it suffices to prove that \\( zlkjhgfd plokijuh asdfghjk plokijuh \\cdots qazxswed plokijuh \\in esfregtb \\) when each \\( qwertyui \\in vfrtgbnh \\). We do this by induction on \\( poiulkjh \\).\n\nFor \\( poiulkjh=1 \\), one sees that \\( zlkjhgfd plokijuh=zlkjhgfd plokijuh \\cdot 1 \\in bnhgvrft \\subseteq esfregtb \\). Now let \\( zlkjhgfd, asdfghjk, \\ldots, jnhtgbrf \\in vfrtgbnh ; zlkjhgfd plokijuh asdfghjk plokijuh \\cdots uytredcv plokijuh= \\) \\( tgbznhyu, jnhtgbrf=rfvtgbyh \\), and \\( tgbznhyu rfvtgbyh plokijuh=yhnujmik \\). Inductively, we assume \\( tgbznhyu \\in esfregtb \\) and seek to show \\( yhnujmik \\in esfregtb \\). The assumption implies that \\( tgbznhyu \\) is in \\( bnhgvrft, cftgvbnh \\), or \\( drtyhnbg \\). If \\( tgbznhyu=mnbvcxzq plokijuh edcrfvtg \\in bnhgvrft \\), then\n\\[\nyhnujmik=(mnbvcxzq plokijuh edcrfvtg) rfvtgbyh plokijuh=mnbvcxzq plokijuh(edcrfvtg rfvtgbyh) plokijuh=mnbvcxzq(edcrfvtg rfvtgbyh)^{-1} plokijuh^{2}(edcrfvtg rfvtgbyh)^{-1} \\in cftgvbnh \\subseteq esfregtb,\n\\]\nusing (i). If \\( tgbznhyu=mnbvcxzq plokijuh^{2} edcrfvtg \\in cftgvbnh \\), then \\( yhnujmik=mnbvcxzq plokijuh^{2} edcrfvtg rfvtgbyh plokijuh \\in drtyhnbg \\subseteq esfregtb \\). If \\( tgbznhyu=mnbvcxzq plokijuh^{2} edcrfvtg plokijuh \\in drtyhnbg \\), then\n\\[\n\\begin{aligned}\nyhnujmik & =mnbvcxzq plokijuh^{2} edcrfvtg(plokijuh rfvtgbyh plokijuh)=mnbvcxzq plokijuh^{2} edcrfvtg\\left(rfvtgbyh^{-1} plokijuh^{2} rfvtgbyh^{-1}\\right) \\\\\n& =mnbvcxzq\\left[plokijuh^{2}\\left(edcrfvtg rfvtgbyh^{-1}\\right) plokijuh^{2}\\right] rfvtgbyh^{-1} \\\\\n& =mnbvcxzq\\left(edcrfvtg rfvtgbyh^{-1}\\right)^{-1} plokijuh\\left(edcrfvtg rfvtgbyh^{-1}\\right)^{-1} rfvtgbyh^{-1} \\in bnhgvrft \\subseteq esfregtb,\n\\end{aligned}\n\\]\nusing (i) and (ii)."
    },
    "kernel_variant": {
      "question": "Let $G$ be a group and let $H\\le G$ be a finite subgroup with \n\\[\n|H|=h<\\infty .\n\\]\n\nAssume that there exist elements $a,b\\in G$ such that \n\\[\na^{2}=1,\\qquad b^{3}=1,\\qquad aba=b^{-1},\\tag{1}\n\\]\nand that both $a$ and $b$ normalise $H$, i.e. \n\\[\naHa^{-1}=H,\\qquad bHb^{-1}=H.\\tag{2}\n\\]\n\n(Thus conjugation by $a$ (resp.\\ $b$) induces automorphisms \n$\\alpha:=\\iota_{a},\\;\\beta:=\\iota_{b}\\in\\operatorname{Aut}(H)$ of orders\n$2$ and $3$.)  \nAssume furthermore that \n\\[\n\\langle a,b\\rangle\\cong{\\mathrm S}_{3},\\tag{3}\n\\]\nso that the relations in (1) are the only ones among $a,b$.\n\nFor the pair $(a,b)$ define the alternating–word set  \n\\[\nP=\\Bigl\\{\nx_{1}t_{1}x_{2}t_{2}\\cdots x_{n}t_{n}\\;:\\;\nn\\ge 1,\\;\nx_{i}\\in H,\\;\nt_{i}\\in\\{a,b\\},\\;\nt_{i+1}\\ne t_{i}\n\\Bigr\\}.\n\\]\n\n(a)  Prove that $P$ is finite.\n\n(b)  Show the sharp upper bound\n\\[\n|P|\\le 6h.\\tag{$\\ast$}\n\\]\n\n(c)  For every prime $p\\ge 5$ construct a quadruple $(G,H,a,b)$\nsatisfying (1)–(3) with\n\\[\n|H|=p^{2}\\quad\\text{and}\\quad |P|=6h,\n\\]\nand hence show that the factor $6$ in $(\\ast)$ is best possible.",
      "solution": "Throughout write  \n\\[\n\\alpha=\\iota_{a}:x\\mapsto axa^{-1},\\qquad\n\\beta =\\iota_{b}:x\\mapsto bxb^{-1}\\qquad(x\\in H),\n\\]\nso that $\\alpha^{2}=\\beta^{3}=1$ and $\\alpha\\beta\\alpha=\\beta^{-1}$\nby (1).\n\n\\medskip\n\\textbf{0.\\; Basic commutation formulas}  \n\nBecause $a^{2}=1$ and $b^{3}=1$, for every $x\\in H$ we have\n\\[\nx\\,a=a\\,\\alpha(x),\\tag{4a}\n\\qquad\nx\\,b=b\\,\\beta^{-1}(x).\\tag{4b}\n\\]\n\n\\medskip\n\\textbf{1.\\; Every alternating word lies in a single $H$–coset}  \n\nPut  \n\\[\nT=\\{1,a,b,b^{2},ab,ba\\}=\\langle a,b\\rangle ,\n\\qquad |T|=6.\n\\]\nWe prove by induction on the length $n$ that every word \n\\[\nW=x_{1}t_{1}\\cdots x_{n}t_{n}\\in P\n\\]\ncan be written  \n\\[\nW=h\\,t,\\qquad h\\in H,\\;t\\in T.\\tag{5}\n\\]\n\n\\emph{Base $n=1$.}  \nIf $t_{1}=a$ then $W=x_{1}a$ and (4a) gives $W=(x_{1})a$ with\n$h=x_{1}\\in H$ and $t=a\\in T$.\nIf $t_{1}=b$ use (4b) analogously.\n\n\\emph{Induction step.}  \nAssume (5) holds for all alternating words of length $n-1$.\nLet\n\\[\nW=U\\,x\\,t_{n},\\qquad \nU=x_{1}t_{1}\\cdots x_{n-1}t_{n-1},\\;\nx\\in H,\\;\nt_{n}\\in\\{a,b\\}.\n\\]\nBy the induction hypothesis choose $h_{0}\\in H,\\;s\\in T$ such that\n$U=h_{0}s$.  Because $s\\in T\\le\\langle a,b\\rangle$ normalises $H$ by (2),\nthe element $x':=sxs^{-1}$ lies in $H$.\n\n\\smallskip\n\\emph{Case $t_{n}=a$.}  Using $x'\\in H$ and (4a):\n\\[\nW=h_{0}sxa=h_{0}x'sa\n      =h_{1}t_{1},\\qquad h_{1}:=h_{0}x'\\in H,\\;t_{1}:=sa\\in T .\n\\]\n\n\\emph{Case $t_{n}=b$.}  Analogously, using (4b):  \n\\[\nW=h_{0}sxb=h_{0}x'sb\n      =h_{2}t_{2},\\qquad h_{2}:=h_{0}x'\\in H,\\;t_{2}:=sb\\in T .\n\\]\n\nThus (5) holds for every $n\\ge 1$.\n\n\\medskip\n\\textbf{2.\\; Bounding the size of $P$}  \n\nFormula (5) implies  \n\\[\nP\\subseteq\\bigcup_{t\\in T}Ht.\\tag{6}\n\\]\nEach left coset $Ht$ has $h$ elements, whence  \n\\[\n|P|\\le \\sum_{t\\in T}|Ht|=6h .\n\\]\nThis proves parts (a) and (b).\n\n(Observe that $H$ itself is among the cosets in (6); for instance\n$baba\\in P$ and $baba=1$ in $\\langle a,b\\rangle\\cong{\\mathrm S}_{3}$.)\n\n\\medskip\n\\textbf{3.\\; Sharpness of the constant $6$}  \n\nFix a prime $p\\ge 5$ and set  \n\\[\nV=\\mathbb{F}_{p}^{2}\\qquad (|V|=p^{2}).\n\\]\nLet  \n\\[\nS=\\begin{pmatrix}0&1\\\\ 1&0\\end{pmatrix},\\qquad\nR=\\begin{pmatrix}0&-1\\\\ 1&-1\\end{pmatrix}\n      \\in{\\mathrm{GL}}_{2}(p).\n\\]\nOne checks\n\\[\nS^{2}=I,\\qquad R^{3}=I,\\qquad SRS=R^{-1},\n\\]\nso that $S,R$ generate a copy of ${\\mathrm S}_{3}$ via $a\\mapsto S,\\;b\\mapsto R$.  \nPut\n\\[\nG:=V\\rtimes\\langle a,b\\rangle\\cong V\\rtimes{\\mathrm S}_{3},\\qquad\nH:=V\\quad(|H|=p^{2}=h).\n\\]\nConditions (1)–(3) are fulfilled, and $\\langle a,b\\rangle\\cap V=\\{1\\}$, so the six cosets $Vt$\n$(t\\in T)$ are pairwise disjoint.\n\nWe now show that every element of each coset $Vt$ lies in $P$, from which \n\\[\n|P|=\\sum_{t\\in T}|Vt|=6|V|=6p^{2}=6h .\n\\]\n\n\\smallskip\n\\textbf{3.1\\; Skeleton words covering the six cosets}  \n\n\\[\n\\begin{array}{lll}\n\\text{coset}&\\text{skeleton word}&\\pi_{\\mathrm{sym}}\\\\[4pt]\nV a      &  v_{1}\\,a                       & a\\\\\nV b      &  v_{1}\\,b                       & b\\\\\nV ab     &  v_{1}\\,a\\,v_{2}\\,b             & ab\\\\\nV ba     &  v_{1}\\,b\\,v_{2}\\,a             & ba\\\\\nV b^{2}  &  v_{1}\\,a\\,v_{2}\\,b\\,v_{3}\\,a   & b^{2}\\\\\nV        &  v_{1}\\,b\\,v_{2}\\,a\\,v_{3}\\,b\\,v_{4}\\,a & 1\n\\end{array}\n\\]\n(here $v_{i}$ are arbitrary elements of $V$).\nEach skeleton is alternating in $a,b$, hence belongs to $P$ for every\nchoice of the $v_{i}$.\n\n\\smallskip\n\\textbf{3.2\\; Surjectivity onto $V$ for each skeleton}  \n\nWrite elements of $G$ multiplicatively as\n\\[\n(v,t)\\,(w,s)=(v+t\\cdot w,\\;ts),\n\\]\nwhere $t\\cdot w$ denotes the natural action of $t\\in{\\mathrm S}_{3}$ on $w\\in V$.\nFor a word $W\\in P$ let $\\pi_{\\mathrm{lin}}(W)\\in V$ be its $V$–component and $\\pi_{\\mathrm{sym}}(W)\\in\\langle a,b\\rangle$ its ${\\mathrm S}_{3}$–component.\n\nFor each skeleton we compute $\\pi_{\\mathrm{lin}}$:\n\n\\begin{enumerate}\n\\item $v_{1}a$: $\\pi_{\\mathrm{lin}}=v_{1}$ – obviously surjective.\n\\item $v_{1}b$: $\\pi_{\\mathrm{lin}}=v_{1}$ – surjective.\n\\item $v_{1}a\\,v_{2}b$:\n      $\\pi_{\\mathrm{lin}}=v_{1}+S v_{2}$ – surjective because $v_{1}$ is free.\n\\item $v_{1}b\\,v_{2}a$:\n      $\\pi_{\\mathrm{lin}}=v_{1}+R v_{2}$ – surjective.\n\\item $v_{1}a\\,v_{2}b\\,v_{3}a$:\n      $\\pi_{\\mathrm{lin}}=v_{1}+S v_{2}+SR v_{3}$ – surjective.\n\\item $v_{1}b\\,v_{2}a\\,v_{3}b\\,v_{4}a$:\n      After the first three letters the group part equals $S$, whence\n      \\[\n      \\pi_{\\mathrm{lin}}\n        =v_{1}+R v_{2}+RS v_{3}+S v_{4},\n      \\]\n      again surjective onto $V$ (already $v_{1}$ alone is arbitrary).\n\\end{enumerate}\n\nHence every coset $Vt$ is completely covered by $P$, and therefore\n\\[\n|P|=6h .\n\\]\n\nSince this construction works for every prime $p\\ge 5$, the constant\n$6$ in $(\\ast)$ cannot be lowered.\n\n\\hfill$\\square$",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.633850",
        "was_fixed": false,
        "difficulty_analysis": "• Two independent generators (a and b) interacting through the non-trivial braid-type relation (★) were introduced; this forces the solver to juggle several simultaneous identities instead of a single one.  \n\n• Alternating words allow the letter to change between a and b at each step, exponentially increasing the number of potential cancellation patterns.  \n\n• The exponent conditions (xa)² = 1 and (xb)³ = 1 lead to different inversion formulas (1)–(2) whose correct, joint use is essential; in the original problem only one such rule was needed.  \n\n• Six double-cosets (not three or two) are required to capture every possible reduction, and each contains h³ elements (rather than h²) because three independent H-factors may survive.  Managing this larger canonical set forces a more intricate induction with many sub-cases.  \n\n• The relation bab = a b⁻¹ a mixes the two generators and must be employed repeatedly to keep the word inside the canonical collection; without it, the previous cancellation tricks do not go through.  \n\nAltogether, the solution demands a significantly broader toolkit: concurrent use of multiple exponent-derived identities, control of mixed generator relations, and a much more elaborate inductive classification, making the variant substantially harder than both the original and the current kernel problems."
      }
    },
    "original_kernel_variant": {
      "question": "Let $G$ be a group and let $H\\le G$ be a finite subgroup with \n\\[\n|H|=h<\\infty .\n\\]\n\nAssume that there exist elements $a,b\\in G$ such that \n\\[\na^{2}=1,\\qquad b^{3}=1,\\qquad aba=b^{-1},\\tag{1}\n\\]\nand that both $a$ and $b$ normalise $H$, i.e. \n\\[\naHa^{-1}=H,\\qquad bHb^{-1}=H.\\tag{2}\n\\]\n\n(Thus conjugation by $a$ (respectively $b$) induces automorphisms \n$\\alpha:=\\iota_{a},\\;\\beta:=\\iota_{b}\\in\\operatorname{Aut}(H)$ of orders\n$2$ and $3$.)  \nAssume furthermore that \n\\[\n\\langle a,b\\rangle\\cong{\\mathrm S}_{3},\\tag{3}\n\\]\nso that the relations in (1) are the only ones among $a,b$.\n\nFor the pair $(a,b)$ define the alternating–word set  \n\\[\nP=\\Bigl\\{\nx_{1}t_{1}x_{2}t_{2}\\cdots x_{n}t_{n}\\;:\\;\nn\\ge 1,\\;\nx_{i}\\in H,\\;\nt_{i}\\in\\{a,b\\},\\;\nt_{i+1}\\ne t_{i}\n\\Bigr\\}.\n\\]\n\n(a)  Prove that $P$ is finite.\n\n(b)  Show the sharp upper bound\n\\[\n|P|\\le 6h.\\tag{$\\ast$}\n\\]\n\n(c)  For every prime $p\\ge 5$ construct a quadruple $(G,H,a,b)$\nsatisfying (1)–(3) with\n\\[\n|H|=p^{2}\\quad\\text{and}\\quad |P|=6h,\n\\]\nand hence show that the factor $6$ in ($\\ast$) is best possible.",
      "solution": "Throughout write  \n\\[\n\\alpha=\\iota_{a}:x\\mapsto axa^{-1},\\qquad\n\\beta =\\iota_{b}:x\\mapsto bxb^{-1}\\qquad(x\\in H),\n\\]\nso that $\\alpha^{2}=\\beta^{3}=1$ and $\\alpha\\beta\\alpha=\\beta^{-1}$\nby (1).\n\n0.\\; Basic commutation formulas  \n\nBecause $a^{2}=1$ and $b^{3}=1$ one has, for every $x\\in H$,\n\\[\nx\\,a=a\\,\\alpha(x),\\tag{4a}\n\\qquad\nx\\,b=b\\,\\beta^{-1}(x).\\tag{4b}\n\\]\n\n──────────────────────────  \n1.\\; Every alternating word lies in a single $H$–coset  \n\nPut  \n\\[\nT=\\{1,a,b,b^{2},ab,ba\\}=\\langle a,b\\rangle ,\n\\qquad |T|=6.\n\\]\nWe prove by induction on the length $n$ that every word \n\\[\nW=x_{1}t_{1}\\cdots x_{n}t_{n}\\in P\n\\]\ncan be written  \n\\[\nW=h\\,t\\qquad\\bigl(h\\in H,\\;t\\in T\\bigr).\\tag{5}\n\\]\n\nBase $n=1$.  \nIf $t_{1}=a$ then $W=x_{1}a$ and (4a) gives $W=(x_{1})a$ with\n$h=x_{1}\\in H,\\;t=a\\in T$.\nIf $t_{1}=b$ use (4b) analogously.\n\nInduction step.  \nAssume (5) holds for all alternating words of length $n-1$.\nLet\n\\[\nW=U\\,x\\,t_{n},\\qquad \nU=x_{1}t_{1}\\cdots x_{n-1}t_{n-1},\\;\nx\\in H,\\;\nt_{n}\\in\\{a,b\\}.\n\\]\nBy the induction hypothesis choose $h_{0}\\in H,\\;s\\in T$ such that\n$U=h_{0}s$.  Since $s\\in T\\le\\langle a,b\\rangle$ normalises $H$ by (2),\nthe element $sxs^{-1}$ lies in $H$.  Write\n\\[\nx' := sxs^{-1}\\in H.\n\\]\n\nNow distinguish the two possibilities for $t_{n}$.\n\nCase $t_{n}=a$.  Using $x' \\in H$ and (4a) we obtain\n\\[\n\\begin{aligned}\nW&=h_{0}sxa=h_{0}(sxs^{-1})sa\n      =h_{0}x'sa        &&(\\text{definition of }x')\\\\\n &=h_{0}x'sa\n      =h_{0}x'(sa)      \\\\\n &=h_{1}\\,t_{1},\\qquad h_{1}:=h_{0}x'\\in H,\\;t_{1}:=sa\\in T .\n\\end{aligned}\n\\]\n\nCase $t_{n}=b$ is completely similar, using (4b):\n\\[\nW=h_{0}sxb\n  =h_{0}(sxs^{-1})sb\n  =h_{2}\\,t_{2},\n\\qquad h_{2}:=h_{0}x'\\in H,\\;t_{2}:=sb\\in T .\n\\]\n\nIn either case $W$ is of the required form (5).\nThus (5) holds for every $n\\ge 1$.\n\n──────────────────────────  \n2.\\; Bounding the size of $P$  \n\nFormula (5) implies the inclusion  \n\\[\nP\\;\\subseteq\\;\\bigcup_{t\\in T}Ht.\\tag{6}\n\\]\nEach left coset $Ht$ has $h$ elements, whence  \n\\[\n|P|\\le \\sum_{t\\in T}|Ht|=6h .\n\\]\nThis proves (a) and (b).\n\n(Observe that $H$ itself is among the cosets in (6); for instance\n$baba\\in P$ and $baba=1$ in $\\langle a,b\\rangle\\cong{\\mathrm S}_{3}$.)\n\n──────────────────────────  \n3.\\; Sharpness of the constant $6$  \n\nFix a prime $p\\ge 5$ and set  \n\\[\nV=\\mathbb{F}_{p}^{2}\\qquad (|V|=p^{2}).\n\\]\nLet  \n\\[\nS=\\begin{pmatrix}0&1\\\\ 1&0\\end{pmatrix},\\qquad\nR=\\begin{pmatrix}0&-1\\\\ 1&-1\\end{pmatrix}\n      \\in{\\mathrm{GL}}_{2}(p).\n\\]\nOne checks\n\\[\nS^{2}=I,\\qquad R^{3}=I,\\qquad SRS=R^{-1},\n\\]\nso that $S,R$ generate a copy of ${\\mathrm S}_{3}$ under $a\\mapsto S,\\;b\\mapsto R$.  \nPut\n\\[\nG:=V\\rtimes\\langle a,b\\rangle\\cong V\\rtimes{\\mathrm S}_{3},\\qquad\nH:=V\\;(=p^{2}\\text{ elements}).\n\\]\nConditions (1)–(3) are fulfilled.\n\nBecause $\\langle a,b\\rangle\\cap V=\\{1\\}$ the six cosets $Vt$\n$(t\\in T)$ are pairwise disjoint.  We now show that each of them is\ncontained in $P$, whence $|P|=6|V|=6p^{2}=6h$.\n\nWrite elements of $G$ multiplicatively as\n\\[\n(v,t)\\,(w,s)=(v+t\\cdot w,\\;ts)\n\\qquad(t,s\\in\\langle a,b\\rangle,\\;v,w\\in V),\n\\]\nwhere $t\\cdot w$ denotes the natural action of ${\\mathrm S}_{3}$ on $V$.\n\nFor a word $W\\in P$ denote by\n$\\pi_{\\mathrm{lin}}(W)\\in V$ its $V$–component and by\n$\\pi_{\\mathrm{sym}}(W)\\in\\langle a,b\\rangle$ its ${\\mathrm S}_{3}$–component.\n\n──────────────────────────  \n3.1\\; Skeleton words covering the six cosets  \n\n\\[\n\\begin{array}{lll}\n\\text{coset}&\\text{skeleton word}&\\pi_{\\mathrm{sym}}\\\\[3pt]\nV a      &  v_{1}\\,a                       & a\\\\\nV b      &  v_{1}\\,b                       & b\\\\\nV ab     &  v_{1}\\,a\\,v_{2}\\,b             & ab\\\\\nV ba     &  v_{1}\\,b\\,v_{2}\\,a             & ba\\\\\nV b^{2}  &  v_{1}\\,a\\,v_{2}\\,b\\,v_{3}\\,a   & b^{2}\\\\\nV        &  v_{1}\\,b\\,v_{2}\\,a\\,v_{3}\\,b\\,v_{4}\\,a & 1\n\\end{array}\n\\]\n(here $v_{i}$ are arbitrary elements of $V$).\nEach skeleton is alternating in $a,b$, hence belongs to $P$ for every\nchoice of the $v_{i}$.\n\n──────────────────────────  \n3.2\\; Surjectivity onto $V$ for each skeleton  \n\n(i) $v_{1}a$: $\\pi_{\\mathrm{lin}}=v_{1}$ – surjective.\n\n(ii) $v_{1}b$: $\\pi_{\\mathrm{lin}}=v_{1}$ – surjective.\n\n(iii) $v_{1}a v_{2}b$:\n$\\pi_{\\mathrm{lin}}=v_{1}+S v_{2}$ – surjective because\n$v_{1}$ is free.\n\n(iv) $v_{1}b v_{2}a$:\n$\\pi_{\\mathrm{lin}}=v_{1}+R v_{2}$ – surjective.\n\n(v) $v_{1}a v_{2}b v_{3}a$:\n$\\pi_{\\mathrm{lin}}=v_{1}+S v_{2}+SR v_{3}$ – surjective\n(same reasoning).\n\n(vi) $v_{1}b v_{2}a v_{3}b v_{4}a$:\n$\\pi_{\\mathrm{lin}}=v_{1}+R v_{2}+RS v_{3}+SR v_{4}$ – surjective.\n\nHence every coset $Vt$ is completely covered by $P$.\nConsequently\n\\[\n|P|=\\sum_{t\\in T}|Vt|=6|V|=6p^{2}=6h,\n\\]\nshowing that the bound $(\\ast)$ is attained.  As there are infinitely\nmany primes $p\\ge 5$, the factor $6$ cannot be lowered.\n\n\\hfill$\\square$",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.504703",
        "was_fixed": false,
        "difficulty_analysis": "• Two independent generators (a and b) interacting through the non-trivial braid-type relation (★) were introduced; this forces the solver to juggle several simultaneous identities instead of a single one.  \n\n• Alternating words allow the letter to change between a and b at each step, exponentially increasing the number of potential cancellation patterns.  \n\n• The exponent conditions (xa)² = 1 and (xb)³ = 1 lead to different inversion formulas (1)–(2) whose correct, joint use is essential; in the original problem only one such rule was needed.  \n\n• Six double-cosets (not three or two) are required to capture every possible reduction, and each contains h³ elements (rather than h²) because three independent H-factors may survive.  Managing this larger canonical set forces a more intricate induction with many sub-cases.  \n\n• The relation bab = a b⁻¹ a mixes the two generators and must be employed repeatedly to keep the word inside the canonical collection; without it, the previous cancellation tricks do not go through.  \n\nAltogether, the solution demands a significantly broader toolkit: concurrent use of multiple exponent-derived identities, control of mixed generator relations, and a much more elaborate inductive classification, making the variant substantially harder than both the original and the current kernel problems."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}