# I want to know the hardness of computing a or b given $g^{b^{-1}}$ and ab in cyclic groups with large prime order

$$G$$ is a multiplicative cyclic group of a large prime order $$p$$ and $$g$$ is a generator of $$G$$

• Theorem 1: Given $$g^{b^{-1}}$$ and $$ab$$, it's hard to compute $$a$$ or $$b$$, where $$a$$ and $$b$$ are randomly picked in $$Z_p$$
• Theorem 2: It's hard to distinguish between the following distributions: $$(g^{b^{-1}},ab)$$ and $$(g^{b^{-1}},z)$$ where $$a$$, $$b$$, and $$z$$ are randomly picked in $$Z_p$$
Is there any proof that these two theorems hold or do not hold?
• Theorem 2 holds trivially. Those distributions are identical. – Maeher Jul 13 '20 at 18:32
• Is this homework? Note that $b^{-1}$ calculated with $\varphi(p)$. So you can consider it as $x$. – kelalaka Jul 13 '20 at 19:25

Theorem 1 is not true in general; consider a group where the discrete log problem is easy. On the other hand, it is true if we assume that the discrete log problem in $$G$$ is hard (that is, if we can recover either $$a$$ or $$b$$ given $$g^{b^{-1}}, ab$$, then we can use that as an Oracle to recover $$x$$ given $$g^x$$)
Theorem 2 is true; for any $$b \ne 0$$ (and we know $$b \ne 0$$ if $$b^{-1}$$ exists), the distributions of $$ab$$ and $$z$$ (for random $$a, z$$) are identical (and hence "hard to distinguish").