MATH.8:5.2 - Binary arrangements on a cycle

Consider binary strings of length four with exactly two entries equal to 1. Positions are numbered 0 through 3 around a cycle. Let r move each entry to the next position, wrapping the last to the first. Four rotations return the original string, and rotation preserves the number of ones.

Starting with 1100, repeated rotation gives:

1100 -> 0110 -> 0011 -> 1001 -> 1100.

There are four distinct members. Only the identity rotation fixes 1100, agreeing with the count 4/1=4.

Starting instead with 1010 gives:

1010 -> 0101 -> 1010.

A rotation by two positions fixes either alternating string. Its stabilizer has two members, so the orbit count is 4/2=2.

Every two-one string either has adjacent ones around the cycle or has opposite ones. This covers the six possible strings and separates the two orbits. If the question asks for arrangements up to rotation, two representatives suffice. If it asks for every labeled string, return all six.

If position 0 receives a distinguished mark that must remain fixed, only the identity rotation preserves that data. The old orbit classification then forgets a distinction required by the new problem. The strings remain available; their identification must change.

Instead put identical marks at positions 0 and 2, with neither mark distinguished from the other. A single rotation and its inverse move the marked set to {1,3}; a half-turn returns it to {0,2}. The data stabilizer is therefore {e,r²}. To find this subgroup, examine compositions such as r² as well as the supplied generators.