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 D_{4}as a subgroup of S_{6}.

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^{-1}s).