# Decrement of a Permutation vs Number of Transpositions

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. I think you could safely say that the decrement gives the minimum possible number of transpositions needed to represent a given permutation. Commented Mar 15, 2023 at 16:04
• @paw88789 That makes sense. How, then, can it be shown that the parity is unique? Perhaps this is its own question though. Commented Mar 15, 2023 at 16:20
• @paw88789: I think that comment should be an answer – I don't think there's much more to say about this question than that. Commented Mar 15, 2023 at 16:23
• @jamman2000: The parity is unique because it's the parity of the permutation. Commented Mar 15, 2023 at 16:25