Thread: Disjoint Cycle Decomposition

Label the vertices of a square 1, 2, 3, 4 in counterclockwise order, with 1 being at the top right. The group D4 acts on the vertices as permutations [e.g., r acts as (1234) and s acts as (24)] and thus also D4 acts as permutations on pairs of vertices: r([1, 2]) = [2, 3], r([1, 3]) = [2, 4], and so on. There are six pairs of vertices: P = {[1, 2], [1, 3], [1, 4], [2, 3], [2, 4], [3, 4]}.

Find the disjoint cycle decomposition of r and s as permutations of the six elements of P.

for notational convenience let us call:

a = {1,2}
b = {1,3}
c = {1,4}
d = {2,3}
e = {2,4}
f = {3,4}

then r induces this permutation on the set P:

a-->d
b-->e
c-->a
d-->f
e-->b
f-->c. in cycle notation this is: (a d f c)(b e) (note that this has order 4, the same as r). if we take s to be (2 4), then acting on P we have for s:

a-->c
b-->b
c-->a
d-->f
e-->e
f-->d. in cycle notation, this is: (a c)(d f).

note that gives an isomorphism of D₄ as a subgroup of S₆.

one can readily verify that (a c)(d f)(a d f c)(b e) = (a f)(b e) = (a c f d)(b e)(a c)(d f) (that is sr = r^-1s).

If I wanted to find the orbits of the action of D4 on P how would I do that?

let's start with the orbit containing a, first.

r takes a to d, so the orbit of a contains d. by inspection, we see that the orbit of a must contain every element of the 4-cycle (a d f c):

r²(a) = f
r³(a) = c

we don't get any new elements of the orbit of a from applying s (so now new elements from applying r or s in any combination), so the orbit containing a is {a,c,d,f}.

that leaves the orbit containing b.

r(b) = e, so the orbit of b contains e. s leaves {b,e} fixed, so there were are:

{{a,c,d,f},{b,e}} is the set of orbits.

thus this action isn't transitive, we have more than one orbit (you can think of it this way: the vertex-pairs {1,2}, {2,3}, {3,4} and {1,4} really "represent the square (sides)" the vertex-pairs {1,3} and {2,4} represent the diagonals (think of the square as having an X inside it). it should be clear that an element of D₄ maps "sides to sides" and "diagonals to diagonals").


if you REALLY want to blow your mind, think about this:

we know D₄ has a normal subgroup, <r>. what does this mean geometrically? well, imagine that we have a square made out of "really stretchy material". if we "mod out the rotations", what we are really doing is identifying all 4 sides,. so imagine "curving the square", blowing it up like a parachute, and then drawing the edges tight (like with a drawstring), so that our square has now become a sphere. our diagonals are now two circles on the sphere: one going from the north pole to the south pole, another at the equator.

the coset <r> is now is the sphere turning on its axis, and the coset s<r> is the sphere turning "end over end" (flipping the poles).