Is no corner twisted or edge flipped by the action of the two squares group $\langle F^2,B^2,U^2,D^2,L^2,R^2\rangle$ on the configurations of the Rubik's cube?
I think the answer is yes, but think the proof would be too long and involves the cycle structure for each of the individual rotations. Any hints? Thanks beforehand.
On
If you by twisted corners or flipped edges mean situations where two configurations have the pieces in the same position, but they differ in orientation of the pieces - this cannot happen in that group.
This is guaranteed by the fact that the operations preserves the axis of the normal vectors of all surfaces, but the normal vector can become opposite. Given the position of a piece this uniquely determines the face normals and by that also the orientation.
In addition you have that the operations so that edges in the $x-y$, $x-z$ and $y-z$ plane remains in that plane. Also corners remain in one of two groups $xyz=\pm 1$.
This makes the group being simultaneuos permutations on five sets of four elements each. I think these are almost independent: the parity of the permutations on the corners have to have the same parity and the parity of the permutations of the edges has to add up to even. This would make it $(4!)^5/4$ elements.
Take a Rubik's cube and mix it up using half turns only. Notice that this only mixes up colors on opposite sides of the cube. For example, on a standard cube the red and orange facelets will be exchanged, but not intermixed with the other four colors. The same with green/blue and white/yellow. It's straight forward to look at the half turns and see that this must be the case.
Now, if a piece is brought back to the same position through some set of half turns, it must preserve the color restriction above. And since no individual piece has colors from opposite faces, (that is, a piece can't have red AND orange), the only twist that preserves the color restriction is the trivial one. Which is to say, pieces cannot be twisted.