Diffie-Hellman should size of group be a prime?

Question

I have found some contradicting answers regarding finding a safe prime.

When p is a "safe prime", this means that $(p−1)/2$ is also prime. We then define $q=(p−1)2$. In that situation, the order of any non-zero $g\mod p$ (except 1 and p−1) is either $q$ or $2q$

If it can produce a group of size $2q$ then the groups length is by definition not a prime.

Okay, so following this theory, 83 should be regarded as a safe prime.

$(83-1)/2=41$ and $41$ is a prime.

The length of the subgroup produced by $2 \mod 83$ is $82$. (can you phrase it like that? Even though it refers to a set/group of congruences by $2^{1..p-1} \mod p$?

I can prove it with a little program

const groupSize = (p, g) => {
    const group = new Set()
    for(let i = BigInt(0); i < p-BigInt(1); i++){
        group.add((g**i)%p)
    }
    console.log(group.size)
}
groupSize(BigInt(83), BigInt(2))

output: 82

This contradicts an answer to one of my previous questions regarding choosing a safe prime.

If h is a factor of the size of the group generated by g, then given $g^x\mod p$, we can compute $h\mod n$ in $O(\sqrt h)$ time. If g generates the entire group, well, its size will be $p−1$, which always has a factor of 2 (assuming $p>2$), and so we'd be giving away $x\mod 2$ for free.

Diffie-Hellman what is the subgroup

This makes me believe that the group cannot be an even number, since it is then a factor of 2?

Answer 1

The length of the subgroup produced by $2\pmod{83}$ is $82$. Can you phrase it like that?

That would be unusual. At least, "length" should be size (cardinality would be very formal). I would use something on the tunes of:

• the order of the multiplicative subgroup generated by $2$ modulo $83$ is $82$.
• $2$ has order $82$, modulo $83$.

---

The previous answer suggests not to choose $g=2$, because it generates the entire multiplicative subgroup $\Bbb Z_{83}^*$, that is the set $[1,83)$ under multiplication modulo $83$, of order $82$. Thus (as explained), with $g=2$, giving $y=g^x\bmod 83$ leaks if $x$ is even or odd: just compute $y^{41}\bmod83$, and that's $1$ if and only if $x$ is even.

Instead, we can use $g=3$, which has order $41$. Or, as pointed in the other answer, $g=4$ which always has prime order if $p$ and $q=(p-1)/2$ are prime.

Answer 2

This contradicts an answer

In what sense do they contradict?

The first one states that, for a safe prime, we have subgroups of size $q$ and $2q$.

The second one states that we typically use a $g$ that generates a subgroup whose size is a prime.

These two statements do not contradict each other; combined, that would mean that (to follow both) we select a $g$ that generates a subgroup of size $q$. This is always possible (in fact, $g=4$ is always such a generator).

Your computations show that, for $p=83$, then $g=2$ is not such a generator. What that means is that, to follow the advise of the second statement, you'd pick a different generator (and, in fact, $g=3$ works, that is, generates a subgroup of size $q$, for $p=83$)