this explains why 1 of 6 will be valid if split gen right?
Yes. But this is due only to the way the programs we use to reassemble the pieces for the final private key work.
Let s (Q = sG) be the secret private key (the public key). Only one person knows 's'.
Then he sends only Q to another person, who run a "split gen" program.
Usually "split gen" works like this:
it computes only (s+1)G = Q + G, (s+2)G = Q + 2G, …, (s+k)G = Q + kG = P
until it gets a public key P with a particular address.
The program knows only k (the partial private key).
Then you need to compute k' (k' G = P):
k' = (s + k) mod n.
Now suppose you get P in this way: - (Q + kG) = P
then in order to get from the partial private key k the correct final private key k' = (-s -k) mod n
you need to know how k was obtained (exploiting the symmetry).
If you get P in this way: λ(Q + kG) = P
(exploiting the endomorphism) the correct final private k' will be: λ*(s + k) mod n and so on.
