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For a permutation group, $c_r(g)$ is the number of length-$r$ cycles of $g$. The cycle index records the average cycle structure. On coloring a finite underlying set, a coloring fixed by $g$ is constant on each cycle. Replacing $t_r$ by the power-sum symmetric polynomial $p_r(x)$ therefore gives the pattern inventory by the weighted Burnside lemma.

For a finite group action $G\curvearrowright X$, Burnside's lemma statesCount pairs $(g,x)$ with $gx=x$ in two ways. Counting first by $g$ gives the displayed sum. Counting first by $x$ gives ${\sum }_x|G_x|$. For an orbit $\mathcal{O}$, the orbit-stabilizer theorem gives $|\mathcal{O}||G_x|=|G|$, so each orbit contributes exactly $|G|$. Division proves the formula.

Use colors $1,\dots ,m$, with variables $x_1,\dots ,x_m$, on the underlying finite set $S$. The weight of a coloring $f:S\to \{1,\dots ,m\}$ is $w(f)={\prod }_{s\in S}{x}_{f(s)}$. It is constant on each coloring orbit. The pattern inventory is $F_G(x)={\sum }_{\mathcal{O}}w(\mathcal{O})$, one weight for each orbit, including its multiplicity when different orbits have the same content.

If $c_r(g)$ is the number of length-$r$ cycles of $g$ on $S$, define the cycle indexIn the symmetric-function cycle-indicator convention, write $Z_G(x)$ for its substitution $t_r=p_r(x)={\sum }_{i=1}^m{x}_i^r$. A fixed coloring must be constant on every cycle of $g$. Coloring one length-$r$ cycle therefore has weight sum $p_r(x)$, independently of other cycles. The weighted Burnside lemma, proved by exactly the same orbit-stabilizer double count with weights, yieldsThis defines both sides in the same alphabet; the raw cycle index in the independent variables $t_r$ is specialized before comparison. The identity holds for all finite $m$ and hence stably for arbitrarily many colors.

For the last assertion, the permutation character on $T$ satisfies $\chi (g)=|{\mathrm{Fix}}_T(g)|$. A pair $(s,t)$ is fixed in the diagonal action on $T\times T$ exactly when both components are fixed. ThereforeThe character is real, so its character inner product with itself and another application of Burnside's lemma giveThis is the rank of a permutation action formula, and in fact does not require transitivity. If the action is transitive and $|T|\ge 2$, the diagonal is one orbit; rank two means exactly one further orbit on distinct ordered pairs, which is double transitivity.