Lemma 1

Let $(\zeta_i)_{i \in I}$ be a family of cycles whose supports are pairwise disjoint. Then the $\zeta_i$ are pairwise permutable. Let $\sigma = \prod_{i \in I} \zeta_i$ and $\bar{\sigma}$ be the subgroup generated by $\sigma$. Then $\sigma(x) = \zeta_i(x)$ for $x \in S_i, i \in I$, and $\sigma(x) = x$ for $x \notin \bigcup_{i \in I} S_i$. The mapping $i \mapsto S_i$ is a bijection of I onto the set of $\bar{\sigma}$-orbits not consisting of a single element.

Let $\zeta$ and $\zeta'$ be two cycles whose supports are disjoint. If

then $\zeta \zeta'(x) = \zeta' \zeta(x) = x$. If $x$ belongs to the support of $\zeta$, then $\zeta'(x) = x$, and $\zeta(x)$ belongs to the support of $\zeta$, whence $\zeta \zeta'(x) = \zeta' \zeta(x) = \zeta(x)$. Similarly when $x$ belongs to the support of $\zeta'$, then $\zeta' \zeta(x) = \zeta \zeta'(x) = \zeta'(x)$. Hence $\zeta \zeta' = \zeta' \zeta$. Therefore the $\zeta_i$ are pairwise permutable and, for $i \in I$ and $x \in S_i$, $\sigma(x) = \zeta_i(x) \in S_i$. The mappings $\sigma$ and $\zeta_i$ coincide on $S_i$, hence $S_i$ is stable under $\sigma$ and the subgroup of $\mathfrak{S}_{S_i}$ generated by the restriction of $\sigma$ to $S_i$ operates transitively on $S_i$; therefore $S_i$ is a $\bar{\sigma}$-orbit. As the $S_i$ are non-empty and are pairwise disjoint, the mapping $i \mapsto S_i$ is injective. As $\bigcup_i S_i$ is the set of $x$ such that $\sigma(x) \neq x$, every $\bar{\sigma}$-orbit not consisting of a single element is one of the $S_i$.