Question

is U 11 is cyclic group​

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

Thm 10 tells us that if G = 〈a〉 is a finite cyclic group of order k, then for any integer m, 〈am〉 = (ad) , where d = gcd (m, k) . So in this exercise, we'll use a = 2 as our generator of (U11, ·) and k = 10 is the order of (U11, ·) . ... The other generators of this subgroup are of the form: 2m, where gcd (m, 10) = 2.