Let $P$ be the finite-support product of the Cohen forcing. It adjoins a sequence of Cohen-generic reals say $a_n$, $n<\omega$, which one naturally calls a $P$-generic sequence. Suppose that $\pi$ is a permutation of $\omega$. Consider the new sequence of reals $a'_n=a_{\pi(n)}$.

Will the new sequence be $P$-generic?

It's answered in the positive provided $\pi$ belongs to the ground model $M$.

Now, what about the case $\pi\notin M$, but $\pi$ belongs to the extension $M[(a_n)_{n<\omega}]$?

• I think the answer is no but for silly reasons. You could choose $\pi$ so that the sequence of first terms of the rearranged Cohen reals, namely $a'_0(0), a'_1(0), a'_2(0), a'_3(0), \dots$, is just the sequence $0,1,0,1,0,1,\dots$ (or whatever other string of 0's and 1's you like, provided you include infinitely many 0's and infinitely many 1's). Do you want to exclude this sort of thing somehow, or does this answer your question? – Will Brian Sep 19 '18 at 9:37
• This was something I thought about when I was finishing my masters degree. I remember talking with Matti Rubin about applying permutations from the generic extensions to the ground model poset. He didn't think it would work. This later turned into my work about iterating symmetric extensions, which is not quite the same thing, of course. Unfortunately, I don't remember what was the outcome of my attempts back at the day. But I think you can in general code bad things into the sequence, like @Will suggested. – Asaf Karagila Sep 20 '18 at 6:18
• I imagine that the application of this sort of thing would be to some kind of choiceless result (perhaps choiceless in the sense of "OD objects" in the ZFC context). So now I'm curious as to what is the ultimate goal here. – Asaf Karagila Sep 20 '18 at 11:43
• My interest is to somehow distinguish those permutations which are generic from those (as in the 0,1,0,1,0,1-example above) which are not. Could it be that just every $\pi$ not in the ground universe produces a non-generic permutation? – Vladimir Kanovei Sep 20 '18 at 13:14
• That is definitely not the case, I think. Take $\pi$ which is in $M[a_0]$, and $\pi(0)=0$. Then by the fact the rest of the sequence is generic over $M[a_0]$, applying it should also produce something generic over that model, so $(a_n')_{n<\omega}$ will also be $M$-generic. (This idea can be extended by quite much, so it will be difficult to give a simple condition as to when the sequence is not generic.) – Asaf Karagila Sep 20 '18 at 14:36