Given a permutation with $n$ items and $c$ independent cycles, the decrement is defined as $n-c$. Here is a simple proof showing that this number is also equal to the number of transpositions in the permutation:
Let $s_i$ be the number of items in the $C^\text{th}$ cycle of the permutation. Then
$$ \sum_{i=1}^cs_i=n. $$
Take as given that a cycle with $s$ items can be reduced into a product of $s-1$ transpositions. Then the total number of transpositions to produce the full permutation is
$$ \sum_{i=1}^c(s_i-1)=\sum_{i=1}^cs_1-\sum_{i=1}^c1=\sum_{i=1}^cs_1-c=n-c $$
We see that the total number of transpositions is equal to the decrement of the permutation, $n-c$.
My question is this:
In the book "Group Theory and its Applications to Physical Problems" by Morton Hamermesh, it states that (chapter 1-2, page 14): "if the decrement of a permutation is even (odd), its resolution into a product of transpositions will have an even (odd) number of factors." I am guessing that if my above proof is correct and the number of factors is equal to the decrement, then this would've been stated in the book rather than a statement about parity. Can someone point out where there is a flaw in my proof? Otherwise, where there may be exceptions to my result such that the decrement does not equal the number of factors, yet parity is preserved?
Perhaps the issue is: Writing a permutation as a product of transpositions is not unique--even the number of transpositions used is not unique. The parity of the number of transpositions is, however, unique. The decrement gives the minimum possible number of transpositions needed to represent a given permutation.
Note: This is a comment converted to an answer as suggested in another comment.