Use of the commutator to reduce the degree of a permutation I've found the following claim here (page 4, in the proof of thm 1.1).
Set $A$ the support of a permutation $\tau$ in $S_n$, with $\deg\tau=|A|=k$.
Given a permutation $\pi$ in $S_n$ such that
$$|A\cap A^\pi| \le 1+(k^2/n)(1+\delta)$$
we have that
$$\deg[\tau,\tau^\pi]\le 3+3(k^2/n)(1+\delta)$$
How can I prove this?
(notation:


*

*the support of a permutation is the set of elements it moves

*$A^\pi$ is the set $\{a^\pi\bigm|a\in A\}$ and $a^\pi$ is the element obtained by applying $\pi$ to $a$

*$\tau^\pi$ is the conjugate of $\tau$ by $\pi$: $\tau^\pi = \pi^{-1}\tau\pi$

*$[\tau,\tau^\pi]$ is the commutator of these two elements: $[\tau,\tau^\pi]=(\tau^{-1}\pi^{-1}\tau^{-1}\pi)(\tau\pi^{-1}\tau\pi)$ )



Edit: this question is a generalization of Commutator is a 3-cycle if...
 A: UPDATE 01/19/2014 : This answer which initially contained a
long comment has now been updated to a full answer.
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To expand on Derek Hold’s comments (which are only partially correct)
Let $U=A\cap A^{\pi}$, $\beta=\pi^{-1}\tau\pi$ and $\gamma=[\tau,\beta]=\tau^{-1}\pi^{-1}\tau^{-1}\pi\tau\pi^{-1}\tau\pi$.
Let $x$ be an integer between $1$ and $n$.
Lemma 1. (1) If $x\not\in A$ and $\pi^{-1}\tau\pi x\not\in A$, then $\gamma x=x$.
(2) If $\pi x\not\in A$ and $\pi\tau x\not\in A$, then $\gamma x=x$.
Proof of lemma 1. (1) Under those hypotheses we have $\tau^{i}(x)=x$
and $\tau^{i}(\pi^{-1}\tau\pi x)=\pi^{-1}\tau\pi x$ for every integer $i$, and hence
$$
\gamma x=\tau^{-1}\pi^{-1}\tau^{-1}\pi\tau\big(\pi^{-1}\tau\pi x\big)=
\tau^{-1}\pi^{-1}\tau^{-1}\pi\big(\pi^{-1}\tau\pi x\big)=\tau^{-1} x=x
$$
(2) Under those hypotheses we have $\tau^{i}(\pi x)=\pi x$
and $\tau^{i}(\pi\tau x)=\pi\tau x$ for every integer $i$, and hence
$$
\gamma x=\tau^{-1}\pi^{-1}\tau^{-1}\pi\tau\pi^{-1}\tau\big(\pi x\big)=
\tau^{-1}\pi^{-1}\tau^{-1}\pi\tau\pi^{-1}\big(\pi x\big)=
\tau^{-1}\pi^{-1}\tau^{-1}\big(\pi\tau x\big)=
\tau^{-1}\pi^{-1}\big(\pi\tau x\big) =x. 
$$
Lemma 2.  If $x$ satisfies neither of the hypotheses in (1) or (2) of the above
lemma, then one of the following three must hold : $\pi x\in U,\pi\tau x\in U,\tau\pi x\in U$.
Proof of lemma 2 If $x$ does not satisfy the hypotheses of (1), then either
$x\in A$ (case 1) or $\pi^{-1}\tau\pi x \in A$ (case 2).
If $x$ does not satisfy the hypotheses of (2), then either
$\pi x\in A$ (case 3) or $\pi\tau x \in A$ (case 4).
Combining the two disjunctions, we have four cases :
Case 1,3 : $x\in A,\pi x\in A$. Then $\pi x\in U$.
Case 1,4 : $x\in A,\pi\tau x\in A$. Then $\pi\tau x\in U$.
Case 2,3 : $ \pi^{-1}\tau\pi x\in A,\pi x\in A$. Then $\tau\pi x\in U$.
Case 2,4 : $\pi^{-1}\tau\pi x\in A,\pi\tau x\in A$. Here two sub-cases must be
distinguished. If $x\in A$, then $\pi x\in U$. If $x\not\in A$, then $\tau\pi x \in U$.
This concludes the proof of lemma 2.
Finally, combining lemma 1 with lemma 2 we obtain $|{\sf supp}(\gamma)| \leq 3|U|$ as
wished.
