On Symmetric group Let $K$ be an infinite set. If $K = A \cup B \cup C$ where $A, B,C$ are disjoint sets and
$|A| = |B| = |C|,$ Prove that $$\operatorname{Sym}(K) = S_A S_B S_A \cup S_B S_A S_B\;,$$ where $$S_A = \{f \in \operatorname{Sym}(K):f(x) = x \text{ for all } x \in A\}\;.$$  
 A: This was proved for countably infinite $K$ by J. D. Dixon, P. M. Neumann, and S. Thomas, Subgroups of small index in infinite symmetric groups, Bull. London Math. Soc. 18 (1986), 560-586; Lemma, p. 580. The extension to arbitrary infinite $K$ was stated by H. D. Macpherson and P. M. Neumann, Subgroups of infinite symmetric groups, J. London Math. Soc. (2) 42 (1990), 64-84; Lemma 2.1, p. 65. Neither of those papers states the lemma in exactly the form you asked about; for their purposes they only needed to say that $\mathrm{Sym}(K)=\langle S_A,S_B\rangle$. The following is copied verbatim from F. Galvin, Generating countable sets of permutations, J. London Math. Soc. (2) 51 (1995), 230-242, where it was used in solving part of Problem 111 (by J. Schreier) from the Scottish Book. 

[P]ermutations are regarded as right operators, and are composed from left to right. [. . .]LEMMA 2.1. Let $E$ be an infinite set. If $E=A\cup B\cup C$ where $A,B,C$ are disjoint sets and $|A|=|B|=|C|$, then $\mathrm{Sym}(E)=S_AS_BS_A\cup S_BS_AS_B$.Proof. Let $\kappa=|E|$. Consider a permutation $\pi\in\mathrm{Sym}(E)$. It is easy to see that $\pi\in S_AS_BS_A$ if (and only if) $|(B\cup C)\setminus A\pi^{-1}|=\kappa$. In particular, $\pi\in S_AS_BS_A$ if $|C\setminus A\pi^{-1}|=\kappa$; similarly, $\pi\in S_BS_AS_B$ if $|C\setminus B\pi^{-1}|=\kappa$. At least one of these alternatives must hold, since $C=(C\setminus A\pi^{-1})\cup(C\setminus B\pi^{-1})$.

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