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The computational Diffie-Hellman problem states that for a cyclic group $G$ of order $p$ and a generator $g$, it is hard to find the value $g^{xy}$ given only $g^x$ and $g^y$ (but easy if either $x$ or $y$ is known); the easiest way being to compute the discrete logarithm of either $g^x$ or $g^y$ and then calculating $(g^x)^y$ = $g^{xy}$.

This is probably just a big misunderstanding of group theory and/or modular operations on my part, but why is it not possible to just calculate the product $g^x \cdot g^y$ as a "regular" group operation (multiplication modulo $p$)? Is this also computationally difficult, or would the result be wrong?

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up vote 13 down vote accepted

$g^x \cdot g^y \;\;\; = \;\;\; (\hspace{.02 in}g\cdot g\cdot g\cdot \ldots$ [$\hspace{.02 in}x$ of them] $\ldots \cdot g\cdot g\cdot g) \: \cdot \: (\hspace{.02 in}g\cdot g\cdot g\cdot \ldots$ [$\hspace{.03 in}y$ of them] $\ldots \cdot g\cdot g\cdot g)$
$= \;\;\; g\cdot g\cdot g\cdot \ldots$ [$\hspace{.02 in}x\hspace{-0.05 in}+\hspace{-0.05 in}y$ of them] $\ldots \cdot g\cdot g\cdot g \;\;\; = \;\;\; g^{\hspace{.02 in}x+y}$

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Back to my highschool math textbooks it is then. Thank you! – lxgr Mar 5 '14 at 21:55

As a add on to the above answer, it would be insecure to use $g^{x+y}$ as key , because both $g^x$ and $g^y$ will be transmitted publicly and by simply eavesdropping one can easily find the required key i.e $g^{x+y}$.

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