# Diffie-Hellmann generator

As far as I understand, there isn't an easy algorithm to find the generator of a cyclic group used in DH Key exchange. Is it true? Given a large prime number like, e.g. 0x614423aaf0d10001f7d7, how can a generator of its group be computed (and checked that's true)?

As far as I understand, there isn't an easy algorithm to find the generator of a cyclic group used in DH Key exchange. Is it true?

No, it is easy, as long as you know the factorization of $p-1$.

A value $1 < g < p$ will be a generator to a prime $p$ iff for every prime factor $q$ of $p-1$, we have $g^{(p-1)/q} \not\equiv 1 \pmod p$

And, there will be enough generators (specifically, $\phi(p-1)$ of them) that just randomly selecting $g$ randomly, and then testing them by the above process is quite doable.

So, isn't factoring $p-1$ a hard problem? Well, it wouldn't be for the 79 bit value you picked, but that's far too small to be used in a Diffie-Hellman exchange, it is generally accepted that you need at least a 2048 bit prime. So, for a value $p-1$ that size, how do we factor it?

Well, what we don't do is select a large random prime $p$ and try to factor it; instead, we build a value $p-1$ with a known factorization, and check that $p$ happens to be prime. One common way of doing this is to search for $p$ such that $(p-1)/2$ is also prime (this is known as a "Sophie Germain prime") and hence $p-1$ has a known factorization.

That said, you said you were doing this for Diffie-Hellman; generally, for DH, you don't want $g$ to be a true generator (as that'll leak, at the very least, the lsbit of your private exponent); instead, we more usually select $g$ such that the order $q$ is a large prime factor of $p-1$; if we know the size of the subgroup we want, we can stir that into the factorization of $p-1$ when we build it; either by:

• Using a Sophie Germain prime (and so $p-1 = 2q$), if $p \equiv 7 \bmod 8$, then $g=2$ will generate the size $q$ subgroup.

• Or by picking a prime $q$ perhaps 256 bits long, and then searching for a (possibly composite) value $r$ such that $p = 2rq + 1$ is also prime; we can find our generator by selecting an arbitrary $h$ and setting $g = h^{2r} \bmod p$; if $g \ne 1$, then that's our generator.

All that said, I expect that you don't know enough number theory to go through that yourself. For decent DH parameters, I send people to this document for precomputed groups; I would recommend the "2048-bit MODP Group" settings.