a) Z12={,,,,,,,,,,,}
Now, No. of elements of order 1=1()
No. of elements of order 2=1()
No. of elements of order 3=2(,)
No. of elements of order 4 =2(,)
No. of elements of order 6=2(,)
No. of elements of order 12=4(,,,)
Divisor d | Number of elements of order d |
Number of integers between 1 and d relatively prime to d |
1 | 1 | 1 |
2 | 1 | 1 |
3 | 2 | 2 |
4 | 2 | 2 |
6 | 2 | 2 |
12 | 4 | 4 |
b)We know that Zn has exactly one subgroup of order d--call it <a>.Then every element of order d also generates the subgroup <a> since every element of order d generates a subgroup of order d and since Zn is cyclic it has exactly one subgroup of order d.So all the elements of order d generates the same subgroup.Now <a> is a cyclic group since subgroup of a cyclic group is cyclic.In a cyclic group if a is a generator, then ak is also a generator of that group if and only if gcd(k,n)=1.Number of such elements is precisely (d).
So the number of elements of order d in Zn is (n)..........(Proved).
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