Unformatted text preview: n have? (HINT: Remember Euler and his φ function.) Solution. We claim there are φ ( n ) generators. First, assume that G = Z /n Z which we may do since the number of generators will not change if we consider an isomorphic group. Now consider k ∈ Z /n Z . If k 6 = 1 divides n , then k added to itself n/k times will be n , so that k has order n/k . Since n/k 6 = n , k is not a generator. This show that the number of generators is at most φ ( n ). Conversely, assume that k is relatively prime to n . Then if k added to itself m times is n , we have km = n and so n must divide m , and hence k has order n . Thus k is a generator and we have that the number of generators is exactly φ ( n ). 1...
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 Spring '09
 DANBERWICK
 Math, Number Theory, Division, Prime number, Cyclic group

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