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Here's a guess: when you ask to compute h ** (a * b), it actually computes h ** (a * b mod q), where $q$ is the order of $g$ in $G1$. If that is the case, well, a random element from $G2$ is not going to be a divisor of $q$ - any element that is the result of a pairing operation will be; however most elements of $G2$ cannot be the result of a pairing ...

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