# 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.

i did check this link
Thankyou

• @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
• @DrLecter ya right i mean that only. can we compute that inverse of 'a' efficiently? is it a cheap operation to get inverse ? – LearningC Jan 22 '14 at 9:47
• @CodesInChaos sorry if my question was not clear. i read your comment in the crypto.stackexchange.com/questions/8925/…. there you told if order of curve is known then getting P from Q=nP is cheap. what i want can also be considered similar to it no? – LearningC Jan 22 '14 at 9:51
• 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
• @CodesInChaos ok fine. thankyou. so the cost of inverse calculation is some what negligible compared to the point multiplication by scalar. ok this is what i wanted to know. thankyou and if you specify some book or site good to refer to know more on these ECc concepts it'll be helpfull. – LearningC Jan 22 '14 at 10:08