Past exam of the options II for mathematics and philosophy course of the University of Oxford / 2013 / B2 / d / i / Solution

A cyclic group of order six has exactly one subgroup of each order $1,2,3,6$, and no others. Since it is abelian, these are also its four conjugacy classes of subgroups. Thus there is one transitive group action of each size $1,2,3,6$.

An action on six points is a disjoint union of its orbits. Its equivalence type is therefore determined by nonnegative integers satisfyingIf $n_6=1$ there is one possibility. If $n_6=0$, the possibilities $n_3=0,1,2$ allow respectively $4,2,1$ choices of $n_2$, with $n_1$ then fixed. HenceThis is counting finite group actions by transitive orbit types. Equivalently, a generator acts as a permutation whose cycles have lengths dividing six, and these are its eight possible cycle types on six points.