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I was going through Zero Knowledge Set Membership and came across the following:

Given $x \in \mathbb{Z}$ and $g$ is the generator of a multiplicative group $\mathbb{G}$ how do we compute $g^\frac{1}{\delta+x}$.

My doubt is, since $\mathbb{G}$ is a multiplicative group, doesn't that require $\frac{1}{\delta + x} \in \mathbb{Z}$. But since $x \in \mathbb{Z}$ is that possible?

Please help me understand this. Thanks in advance.

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    $\begingroup$ $g^\frac{1}{\delta+x} = g^{{(\delta+x)}^{-1}}$. $\endgroup$ – kelalaka Apr 22 at 10:19
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    $\begingroup$ The division is modulo the order of the group (which is hopefully prime). $\endgroup$ – fkraiem Apr 22 at 10:41
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    $\begingroup$ To be more explicit, in this context $\frac{1}{\delta+x}$ stands for that number $y$ such that $y \times (\delta + x) = 1$ modulo the order of $g$ $\endgroup$ – poncho Apr 22 at 11:30

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