The given permutation are σ=(123) and τ=(12) .
∴σ(τ)=(123)(12)=(13) .
1. The given group is Z6 under addition madulo 6.
We known that Z6 is a cyclic group of order 6 generated by 1 .
i,e,Z6=<1>.
So, O(2)= order of 2 =12=(1+1)=gcd(2,O(1)O(1)=gcd(2,6)6 .
∴O(2)=3 .
2. Let the given ring is
R=<x2+x+1>Z2[x]∴ R={f(x)+<x2+x+1>:f(x)∈Z2[x]}
={ax+b+<x2+x+1>:ax+b∈Z2[x]}
={0+<x2+x+1>,x+<x2+x+1>,x+1+<x2+x+1>,1+<x2+x+1>}
3. \ Let G be a finite group and a∈G such that O(a)=20= order of a .
∴ <a> is a cyclic subgroup of G ,whose order is 20 and a6∈<a>.
Hence ,O(a6)=gcd(6,O(a)O(a)=gcd(6,20)20=220=10.
Where O(a6) represent oder of a6 .
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