Is n/d the Smallest Order for x^k in Group <x>?

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SUMMARY

The discussion centers on proving that the order of the element x^k in a group of order n is n/d, where d is the greatest common divisor (gcd) of n and k. The participant establishes that (x^k)^(n/d) = 1, indicating that n/d is a valid order. However, they seek to confirm that n/d is indeed the smallest order, exploring the implications of assuming a smaller order s < n/d and its relationship to n/d.

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Homework Statement


A group <x> has order n. k= nq+r where 0<= r < n. Prove that the order of x^k is n/d where d = gcd (n,k)


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The Attempt at a Solution



I know that (x^k)^(n/d) = 1, but how do I prove that n/d is the smallest one? I tried to assume that (x^k)^(n/d-q) = 1 but could not arrive at any contradiction.

Thank you!
 
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Let's say x^s = 1, where s < n/d. What must be the relationship between s and n/d?
 

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