Do they mean find a general formula such that for any n we can find its order by substituting its value in the formula? If so how do we go about doing this?
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The additive order of $n-1$ in $\Bbb Z_n$ is $n$. That's it's a generator for the additive group, since $(n,n-1)=1$.
Note: $$n-1=(n-1)\cdot 1\overset{\text{in multiplicative notation}}{\overbrace{=}}1^{n-1}\implies \lvert n-1\rvert =n/(n,n-1)=n.$$
Here's a proof of that formula.
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I dont think they mean for any $n$ since no such formulas exist. For $n-1$, note that the order is just $2$ since $(n-1)^2 \equiv (-1)^2 \equiv 1 \pmod n$.
For addition, the order is infinity for any element in the set since we can never get back to our identity element.
Sandeep Silwal
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Your claim that $n-1$ has order $2$ mod $n$ does in fact hold for all $n$ except $n=2$. – anon Apr 23 '14 at 22:16
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Why do you think that the additive order is infinity? – Bill Dubuque Apr 23 '14 at 22:25
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We are talking about $\mathbb Z_n=\mathbb Z/n\mathbb Z$ – PNM Dec 17 '22 at 07:45