Question about discrete log - why can one attacker not solve for g but two colluders can?

Question

I'm reading about the broadcast encryption described in Fiat and Naor's paper, I screen grabbed the relevant part below. My question is essentially:

If there's a secret:

$g^{abc}$ $mod$ $A$ (where $A=PQ$ (both sufficiently large primes))

why is it that one user who knows:

$g^d$ $mod$ $A$, $a$, $b$, $c$, $d$, and $A$

could not calculate $g^{abc}$ $mod$ $A$ but when you add a second user who know $g^e$ $mod$ $A$, and $e$, they can then collude to calculate $g^{abc}$ $mod$ $A$ ?

Context is provided below.

The full paper can be found here

The paper states that one user not in the privileged set could not calculate the secret, but two colluding users could (and so they rate the scheme 1-resilient).

Answer 1

The idea is that someone who just knows $g^d \bmod N$ and $d$, they cannot compute $g$ (because they don't know the factorization of $N$; hence this is the RSA problem); however, if they know $g^d \bmod N$ and $d$ and $g^e \bmod N$ and $e$, then (assuming $gcd(d, e) = 1$), they can (and once they have $g$, they can compute $g^{abc} \bmod N$).

That's because, since they know $d$ and $e$, they can compute two integers $s$ and $t$ such that $sd + te = 1$ (that's where the assumption that $gcd(d, e) = 1$ comes in; if that's not true, then no such $s$, $t$ will exist). Once that have those two integers, they can compute $(g^d)^s \times (g^e)^t = g^{sd + te} = g^1 = g$, recovering $g$. One of $s$ and $t$ will be negative; that's not an issue, as we can compute inverses modulo a number of unknown factorization.