Multiplicative order of a number

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SUMMARY

The discussion centers on proving that if the multiplicative order of a number \( a \) modulo \( n \) raised to \( s \) equals the multiplicative order of \( a \) modulo \( n \), then \( s \) and the order of \( a \) modulo \( n \) are coprime. The participants explore proof techniques, particularly proof by contradiction, and analyze the implications of the relationship between \( s \) and the order of \( a \). The key conclusion is that if \( (s, \mathop{\mathrm{ord}_n} a) = d \), then \( a^s = (a^d)^e \) leads to further insights about \( \mathop{\mathrm{ord}_n} a^d \).

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koolkuzz
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I'm stuck on a question that requires me to prove the following:

Let s,n ∈ N, a ∈ Z and (a,n) = 1. (Note: a & n are coprime)
Prove that ordnas = ordna implies that (s, ordn a) = 1.

I have tried using Proof by contradiction, but seem to go nowhere with this.

Can you help?
 
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Suppose (s, \mathop{\mathrm{ord}_n} a) = d and put s = de. Observe that a^s = (a^d)^e. What is \mathop{\mathrm{ord}_n} a^d?
 
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