TITLE: Prove or disprove: The minimum number of transpositions needed to decompose $\sigma$ is $n-S$.
QUESTION [10 upvotes]: Assume that $\sigma \in S_n$ and $\sigma = \alpha_1 \dots\alpha_s$ is the decomposition of $\sigma$ into disjoint cycles, such that all of the members of $\{1,2,\dots,n\}$ are appeared in the members of $\alpha_1, \dots, \alpha_s$.   
Is this statement true? 

The minimum number of transpositions needed to decompose $\sigma$ is $n-S$.

REPLY [11 votes]: amakelov explains why $n-S$ transpositions are sufficient. To show that $n-S$ transpositions are necessary, one proves that for a permutation $\sigma$ and a transposition $\tau$, $\sigma\tau$ has one more, or one fewer cycle than $\sigma$. As the identity permutation has $n$ cycles, a product of $k$ transpositions has at least $n-k$ cycles. So if $k<n-S$,
a product of $k$ transpositions has more than $S$ cycles.