// Numbas version: exam_results_page_options {"name": "Conjugate of permutations", "extensions": ["permutations"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"functions": {"orderpossible": {"definition": "var factors = Numbas.math.factorise(n);\nvar t = 0;\nfor(var i=0;i1) {\n out += '^{'+factors[i]+'}';\n }\n }\n}\nreturn out;", "type": "string", "language": "javascript", "parameters": [["n", "number"]]}, "sum": {"definition": "var t = 0;\nfor(var i=0;i0) {\n var cycle_length = Math.pow(Numbas.math.primes[i],factors[i]);\n for(var j=0;j1) {\n out += ' and ';\n } else if(i>0) {\n out += ', ';\n }\n out += '$'+l[i]+'$';\n}\nreturn out;", "type": "string", "language": "javascript", "parameters": [["l", "list"]]}, "permfrompartition": {"definition": "var to = [];\nvar offset = 0;\nfor(var i=0; i0) {\n out += ' + ';\n }\n out += l[i]\n t += l[i];\n}\nout += ' = '+t;\nreturn out;", "type": "string", "language": "javascript", "parameters": [["l", "list"]]}, "randompartition": {"definition": "var l = [];\nwhile(n) {\n var t = Numbas.math.randomint(n)+1;\n l.push(t);\n n -= t;\n}\nreturn l;", "type": "list", "language": "javascript", "parameters": [["n", "number"]]}, "tracemap": {"definition": "n -= 1;\nvar l = perms.length;\nvar out = ''+(n+1);\nfor(var i=l-1;i>=0;i--) {\n n = perms[i].map(n);\n out += ' \\\\to '+(n+1);\n}\nreturn out;", "type": "string", "language": "javascript", "parameters": [["n", "number"], ["perms", "list"]]}}, "name": "Conjugate of permutations", "tags": ["permutations"], "type": "question", "advice": "

You could find $\\tau \\sigma \\tau^{-1}$ by first finding $\\tau^{-1}$ and then performing the composition by hand, but there's a quicker way: $\\tau \\sigma \\tau^{-1}$ has the same cycle structure as $\\sigma$, but each $n$ is replaced by $\\tau(n)$.

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Proof
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Assume, without loss of generality, that $\\sigma$ is a single cycle $(b_1, \\dots, b_n)$, and let $\\pi = \\tau \\sigma \\tau^{-1}$. Look at what happens to $\\tau(b_i)$ when acted on by $\\sigma$:

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\\[ \\tau(b_i) \\overset{\\tau^{-1}}\\mapsto b_i \\overset{\\sigma}\\mapsto b_{i+1} \\overset{\\tau}\\mapsto \\tau(b_{i+1}) \\]

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So, for any $1 \\lt i \\leq n$, $\\tau(b_i) \\overset{\\pi}\\mapsto \\tau(b_{i+1})$.

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For any $x \\notin \\{b_1,\\dots,b_n\\}$, $\\sigma(x) = x$, so $\\pi(x) = x$.

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Hence, $\\pi$ is the cycle $(\\tau(b_1), \\dots, \\tau(b_n))$.

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a)

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$\\var{conjugate1}$

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b)

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$\\var{conjugate2}$

\n", "rulesets": {}, "parts": [{"prompt": "

Where $\\sigma = \\var{sigma1}$ and $\\tau = \\var{tau1}$, write down the permutation $\\tau \\sigma \\tau^{-1}$.

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$\\tau \\sigma \\tau^{-1}$ = [[0]]\n

", "marks": 0, "gaps": [{"marks": 1, "showCorrectAnswer": true, "displayAnswer": "{show(conjugate1)}", "scripts": {"validate": {"order": "instead", "script": "var Permutation = Numbas.extensions.permutations.Permutation;\ntry{\n Permutation.parse(this.studentAnswer);\n} catch(e) {\n this.giveWarning(\"Invalid notation\");\n return false;\n}\nreturn true;"}, "mark": {"order": "instead", "script": "var Permutation = Numbas.extensions.permutations.Permutation;\ntry {\n var p = Permutation.fromString(this.studentAnswer);\n} catch(e) {\n this.setCredit(0,\"Invalid notation\");\n return;\n}\nif(!Permutation.is_disjoint(this.studentAnswer)) {\n this.setCredit(0,\"Your cycles are not disjoint\");\n return;\n}\nvar unwrapValue = Numbas.jme.unwrapValue;\nvar prod = unwrapValue(question.scope.variables.conjugate1);\nif(p.equal(prod)) {\n this.setCredit(1,\"Correct!\");\n} else {\n this.setCredit(0,\"Wrong\");\n}"}}, "answer": "", "type": "patternmatch"}], "showCorrectAnswer": true, "scripts": {}, "type": "gapfill"}, {"prompt": "

Where $\\sigma = \\var{sigma2}$ and $\\tau = \\var{tau2}$, write down the permutation $\\tau \\sigma \\tau^{-1}$.

\n

$\\tau \\sigma \\tau^{-1}$ = [[0]]\n

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