On creating a single-cycle permutation from a shuffle:
Take 0, 1, 2, 3, 4, 5 and shuffle it. 4, 3, 0, 5, 1, 2
Now you can read out a permutation:
4 → 3
3 → 0
0 → 5
5 → 1
1 → 2
2 → 4
It will only have one cycle.
Take 0, 1, 2, 3, 4, 5 and shuffle it. 4, 3, 0, 5, 1, 2
Now you can read out a permutation:
4 → 3
3 → 0
0 → 5
5 → 1
1 → 2
2 → 4
It will only have one cycle.
If by 'shuffle' you mean like Fisher-Yates, this can't work all the time, since that can produce, e.g. 0, 1, 2, 3, 4, 5 -> 1, 0, 4, 5, 3, which has two cycles. If by 'shuffle' you mean like Fisher-Yates, but not leaving an element in its original position, (i.e., generating a derangement) then this doesn't suffice either, since notice that the same example satisfies that condition and yet still has two cycles. You need Sattolo's algorithm (or something like it) to guarantee a single cycle.
2, 4, 3, 0, 5, 1
you get the same single-cycle permutation.
I think this might actually lead to a easier proof than in the article.