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Group theory

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If G is a cyclic group of order n and suppose a$\in$G is a generator of the group then b=a$^{k}$ and order of b = n/d, where d=gcd(k,n)

so coming to your question n=10, k=8, b=a$^{8}$ order of b = n/d

d = gcd(10,8) =2

therefor order of b is 10/2 = 5
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