In the book [1] on page 95 the following lemma stated:
Lemma. If $\tau$ is an transposition in $\mathscr{S}_n$ and $\phi \in \mathscr{S}_n$ then $ N(\tau \phi) = N(\phi) \pm 1$
where $N(\sigma)$ is the number of orbits of $\sigma \in \mathscr{S}_n$
and a rather lengthy proof (about 2 pages and still incomplete - some cases are left to the reader as exercises) is given. But it seems that a much shorter proof of the lemma is possible.
Note that every nontrivial cycle C is usually represented by a sequence X - a linearization of the cyclic diagram [1] of C. The sequence X can be called a cyclic sequence of the cycle C. Note that the length of $X > 1$.
The main idea of the new proof is to allow in the factorization of a permutation using for every trivial orbit {a} the 1-sequence (a). We may assume that the 1-sequence (a) represents the identity permutation on the set $A_n$.
So every permutation $\sigma$ can be represented as $\sigma = X_1 ... X_k $ where k is the number of orbits of $\sigma$ and the sequence $X_i$ represents a cycle of $\sigma$. It can be called the extended factorization of $\sigma$.
Lemma 1. Let $a \in A_n$, $C$ and $D$ are injective sequences over $A_n, im(C) \cap im(D) = \emptyset, a \notin C,D$. Then $aC \circ aD = aDC$.
Case 1. $C$ is the empty sequence: $aC = (a)$ = the identity permutation, $aD = aD$;
Case 2. $D$ is the empty sequence: $aD = (a) =$ the identity permutation, $aC = aC$;
Case 3. $C$ and $D$ are both nonempty: check $(aC \circ aD)(x) = aDC(x)$ for all $x \in A_n$;
Case 3.1. $x = a : (aC \circ aD)(a) = first(D); aDC(a) = first(D)$;
Case 3.2. $x \in im(D)$ and $x \neq last(D): (aC \circ aD)(x) = next(D,x); aDC(x) = next(D,x)$;
Case 3.3. $x = last(D); (aC \circ aD)(last(D)) = first(C); aDC(x) = first(C)$;
Case 3.4. $x \in im(C)$ and $x \neq last(C): (aC \circ aD)(x) = next(C,x); aDC(x) = next(C,x)$;
Case 3.5. $x = last(C); (aC \circ aD)(last(C)) = a; aDC(last(C)) = a$
Case 3.6. $x \neq a, x \notin im(C), x \notin im(D): (aC \circ aD)(x) = x; aDC(x) = x$;
(end of proof Lemma 1).
The new proof of the lemma from [1]
Let $ \phi = X_1 ... X_k $ be the extended factorization of $\phi$ and let $\tau = (ab)$.
Case 1. $a \in im(X_i), b \in im(X_j), i \neq j$. Then $X_i$ and $X_j$ can be written as $aC$ and $bD$, where $C$ and $D$ are some sequences, possibly empty.
$ab \circ aC \circ bD =$ (by Lemma 1)
$aCb \circ bD =$ (by rotating $aCb$)
$baC \circ bD =$ (by Lemma 1)
$bDaC =$ (by rotating $bDaC$)
$aCbD$
So the number of orbits in this case decreases by 1. Also note that the equation $ab \circ aC \circ bD = aCbD$ has been proved.
Case 2. $a$ and $b$ are both in some $X_i$. Then $X_i$ can be written as $aCbD$, where $C$ and $D$ are some sequences, possibly empty. Multiplying both parts of the equation proved in Case 1 by (ab), we receive $ab \circ ab \circ aC \circ bD = ab \circ aCbD$, which is equivalent to $ab \circ aCbD = aC \circ bD$ (because $ab \circ ab$ is the identity permutation). So the number of orbits, in this case, increases by 1.
[1] Joseph Landin. An Introduction to Algebraic Structures. Dover Publications, New York, 1989.