# inverse of scalar multiplier in ECC

I am learning to use ECC. i got into situation where i have $Q=abG$, where $G$ is the generator of the finite field formed on EC using a prime $p$ modulus and $a$ , $b$ are random numbers. now suppose i know $Q$, $G$ and $a$ , calculating $bG=\text{inverse}(a)Q$ is efficient ? i know the order of the group too. i want to know whether this inverse calculation is cheap enough to be calculated as frequently as random number changes?

i tried to get material for this kind of inverse calculation but dint get as such. If anyone could provide some source to this information on ECC it'll be helpfull.

@Rohith I suppose that $G$ and $Q$ are points on the curve and you mean that $G$ is a generator of your elliptic curve group? Then to compute $bG$ you simply compute $a^{-1}$ as the multiplicative inverse of $a$ in $Z_p$ and compute $bG=a^{-1}Q$ using your scalar multiplier. – DrLecter Jan 22 '14 at 9:43
My mistake. You're right. A simple modular inverse should do the trick. Computing a modular inverse costs about 10% of a scalar multiplication $aQ$. So computing $a^{-1}Q$ is dominated by the cost of multiplying $Q$ with $a$. – CodesInChaos Jan 22 '14 at 9:55