Given a prime $P$ with $$P= r \cdot q+1$$ with $q$ prime as well.
I'm looking for a generator $g$ of the Schnorr subgroup with order $q$ which is small by value and has a inverse (to $\bmod P$) which is small as well: $$|\{g^i, \forall i \in [0,P-1]\}| = q$$ $$g^q = 1 \bmod P$$ $$g < m \text{ } \land \text{ } g^{-1}<m$$

To find a Schnorr group generator $g$ in general we can pick a random value $h$ until $$ h^r \not= 1 \bmod P$$ $$\rightarrow \text{ } g = h^r \bmod P$$ To find $g^{-1}$ we can use the Extended Euclidean algorithm or just compute: $$ g^{-1} = g^{q-1} \bmod P$$


Q1.) What is the best way to find $g,g^{-1}$ which are each less than value $m$? (edit: found some with step-by-step computation. Main interest now is Q3 (edit-end))
Q2.) How small can $m$ be?
Q3.) Would a small $g,g^{-1}$ have any impact to security?

Trial for Q1: Ways to find small $g, g^{-1}$
a.) pick random $v$ number and project to Schnorr group $v_s = v^r \bmod P$
(all values but $1$ of a Schnorr group are generators)
b.) compute inverse with EEA or with $v_s^{q-1} \bmod P$ (any way faster?)
c.) check if $g, g^{-1} < m$ . If not repeat a.)b.)c.)

Another way than random picks all the time would be an initial random pick for $g$, and a 2nd random value $w$ in this group to start from. Compute the inverse of both as well and traverse through the group with new generator $g^*_j$: $$g_j^* = w \cdot (g)^j \bmod P$$ $${g_j^*}^{-1} = w^{-1} \cdot (g^{-1})^j \bmod P$$ $$w \cdot (g)^j \cdot w^{-1} \cdot (g^{-1})^j = 1 \bmod P$$ and test if $g^*_j$ and its inverse are smaller than $m$.
This computes the values faster but will it be faster in finding suitable generator? Can it be for certain values the results lay in a valley of bad values?

Trial for Q2: How small can $m$ be?
The product of $g, g^{-1}$ need to be at least $p+1$
That means $m \ge \sqrt{p-1}$
Is there another higher border for smallest $m$?

In general the equation need to hold: $$g \cdot g^{-1} = a\cdot P +1 \equiv 1 \bmod P $$ $$a \in [1,P-2]$$

Can the factorization of $(aP+1)$ be used to solve Q1 ($g,g^{-1}$ products of factor(combinations))?
Any way to guarantee its inside the Schnorr group?
Or just check again with $g^q = 1 \bmod P$?

  • 1
    $\begingroup$ As for Q2: it's unlikely that you could give significantly larger bounds on $m$; one small example I found was $p=229\cdot 698512 + 1, g = 12645, g^{-1} = 12650, ord(g) = 229$; this example has an $m=12650$ which is not much larger than $\sqrt{p-1} = 12647.4997...$. Obviously, that's much smaller than a cryptographically relevant example, however the existence of such suggests that there are likely to be much larger ones. $\endgroup$
    – poncho
    Feb 3, 2020 at 16:37
  • $\begingroup$ @poncho also did some experiments. It tried with a prime ~69 bit. In first trial I didn't found any factor below $2^{59}$. After changing prime and factors a little (about same size) I could find some small generators (54 and 57 bit). Those are not that small as yours but they are sufficient for me. I wrote some program in C which found them quite fast (like my trial Q2 above (part II)). Now my main interest is Q3, If such small generators with small inverse have an impact to security. $\endgroup$
    – J. Doe
    Feb 3, 2020 at 17:42

1 Answer 1


Q3.) Would a small $g, g^{-1}$ have any impact to security?

It would not appear so. It is easy to show that, if we had a generator $g$ that makes the Discrete Log (or the Computational Diffie-Hellman) problem easy, then the Discrete Log/CDH problem is easy with respect to any base $h$ that's within the same subgroup.

Suppose we had a way that, given $g, g^x$, gave us the value $x$. Then, what we could do, given $h, h^y$ is first find the value $z$ such that $g^z = h$ (by solving the discrete log to the base $g$), and then the value $w$ such that $g^w = h^y$ (again, solving another discrete log to the base $g$); then, we have $h^{wz^{-1}} = h^y$; since we know $w$ and $z$, that tells us the discrete log. Similar (but more complicated) logic allows us to solve the Computational Diffie-Hellman problem, given a CDH oracle with respect to base $g$.

However, this is not a complete answer to the question; it might be that the existence of a small $g, g^{-1}$ makes all discrete log/CDH problems easy.

I don't know of a way this can be disproven. However:

  • None of the known discrete log methods have significant speed-ups when given a small $g, g^{-1}$ (by significant, I mean more than speeding the operation of multiplying by $g$)

  • For just about any prime, there will be tiny $g$ values that generate groups that contain the Schnorr group as a subgroup. If there were a fast way to complete discrete logs with a tiny $g$ in this group, then you could use that to compute discrete logs in the Schnorr group (and although this argument doesn't take into account the tiny $g^{-1}$, there's no obvious way to use that).


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