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(15) Let n E Z+,and let d be a positive divisor of n. Theorem 23.7 tells us that Zn contains exactly one subgroup of order d,Part 15A and 15B

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a) Z12={\bar{0},\bar{1},\bar{2},\bar{3},\bar{4},\bar{5},\bar{6},\bar{7},\bar{8},\bar{9},10,\bar{11}}

Now, No. of elements of order 1=1(\bar{0})

No. of elements of order 2=1(\bar{6})

No. of elements of order 3=2(\bar{4},\bar{8})

No. of elements of order 4 =2(\bar{3},\bar{9})

No. of elements of order 6=2(\bar{2},10)

No. of elements of order 12=4(\bar{1},\bar{5},\bar{7},\bar{11})

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 \phi(d).

So the number of elements of order d in Zn is \phi(n)..........(Proved).

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