Orthogonal generators of a group in Lelantus protocol

Question

In the Lelantus Paper, the authors mentionned this:

In our case, the commitment key ck specifies a prime-order group G and three orthogonal group generators $g, h_1$ and $h_2$.

G is mentioned in the performance section of the paper to be the famous elliptic curve secp256k1. Hence, I don't understand the notion of "group generators" that are orthogonal.

Indeed the group G is of prime-order and obviously commutative (even in the general case). Hence, non-trivial generators are of prime order p as well, and their generated subgroups are the whole group G.

Am I missing something here or is the "orthogonal" notion useless or even wrong?

Note: For me, it would make sense if the sentence meant that $\langle g,\ h_1,\ h_2\rangle\ = \ G$ and the subgroups $\langle g\rangle, \langle h_1\rangle, \langle h_2\rangle$ have trivial intersections together, but the prime order of the group $G$ forbids that, no?

Answer 1

A GUESS: the author meant that three numbers $a,b,c \in \mathbb{Z}$ such that $$g^a\cdot h_1^n \cdot h_2^c =1$$ are infeasible to compute. While the term "orthogonal" seems inappropriate, This is a fairly standard assumption.