The given permutation are "\\sigma=(123) \\ and \\ \\tau=(12)" .
"\\therefore \\sigma(\\tau)=(123)(12)=(13)" .
1. The given group is "\\Z_6" under addition madulo 6.
We known that "\\Z_6" is a cyclic group of order 6 generated by 1 .
i,e,"\\Z_6=<1>."
"So, \\ O(2)=" order of 2 "=1^2=(1+1)=\\frac{O(1)}{gcd(2,O(1)}=\\frac{6}{gcd(2,6)}" .
"\\therefore O(2)=3" .
2. Let the given ring is
"R=\\frac{\\Z_2[x]}{<x^2+x+1>}""\\therefore \\ R=\\{f(x)+<x^2+x+1>:f(x)\\in\\Z_2[x]\\}"
"=\\{ax+b+<x^2+x+1>:ax+b\\in\\Z_2[x]\\}"
"=\\{ 0+<x^2+x+1>,x+<x^2+x+1>,x+1+<x^2+x+1>,1+<x^2+x+1>\\}"
"3. \\" "Let \\ G" be a finite group and "a\\in G" such that "O(a)=20=" order of "a" .
"\\therefore \\ <a>" is a cyclic subgroup of "G" ,whose order is 20 and "a^6\\in<a>."
Hence ,"O(a^6)=\\frac{O(a)}{gcd(6,O(a)}=\\frac{20}{gcd(6,20)}=\\frac{20}{2}=10."
Where "O(a^6)" represent oder of "a^6" .
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