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Notice that the problem is symmetric in $f$ and $g$ so it's enough to prove only one direction. Now suppose $f(u) \in S$ and $g(u) \in T$. We have $fff(u)=ggf(u)$ and $fgg(u)=gfg(u)$ and we consider two cases. i) $u\in S\Rightarrow ff(u)=gg(u)$. Therefore $gfg(u)=fgg(u)=fff(u)=ggf(u)$, which implies $fg(u)=gf(u) \Rightarrow u \in T$ ii) $u\in T \Rightarrow fg(u)=gf(u)$. Therefore $fgg(u)=gfg(u)=ggf(u)=fff(u)$, which implies $gg(u)=ff(u) \Rightarrow u \in S$. So $u \in S\cap T$. By the symmetry of $f$ and $g$ we have the following results 1) $f(u)\in S$ and $g(u)\in T \Rightarrow u \in S\cap T$ 2) $f(u)\in T$ and $g(u)\in S \Rightarrow u \in S\cap T$. Now let $u=f(w)$. If $f(w) \in S\cap T$, then since $g(w) \in V = S\cup T$, we can use either (1) or (2) to yield $w \in S\cap T$. But since $f$ is bijective and $S\cap T$ is finite it is equivalent to $w\in S\cap T\Rightarrow f(w) \in S\cap T$. And by symmetry $w \in S\cap T \Rightarrow g(w) \in S\cap T$. But now we are done because if $f(u) \in S$ then either $g(u) \in S$, which satisfies the condition, or $g(u) \in T \Rightarrow u \in S\cap T \Rightarrow g(u) \in S\cap T$. So $f(u) \in S \Rightarrow g(u) \in S$. Finally, by symmetry $f(u) \in S \Leftrightarrow g(u) \in S\qquad \square$

lomos_lupin wrote: let $ V$ be a finitive set and $ g$ and $ f$ be two injective surjective functions from $ V$to$ V$.let $ T$ and $ S$ be two sets such that they are defined as following" $ S = \{w \in V: f(f(w)) = g(g(w))\}$ $ T = \{w \in V: f(g(w)) = g(f(w))\}$ we know that $ S \cup T = V$, prove: for each $ w \in V : f(w) \in S$ if and only if $ g(w) \in S$ Claim : If $f(x) \in S, g(x) \in T$, then $x \in S \cap T$. Proof. Note that $f^3 = g^2f$ and $gfg = fg^2$. If $x \in S$, then $gfg = fg^2 = f^3 = g^2f$ which implies $x \in T$. If $x \in T$, then $f^3 = g^2f = fgf = fg^2$ which implies $x \in S$. Since $V = S \cup T$, the desired result holds true. $\square$ Claim 2 : If $f(x) \in S \cap T$, $x \in S \cap T$ and $g(x) \in S \cap T$ are equivalent statements. Proof. If $g(x) \in S$, then by Claim 1, $x \in S \cap T$. Similar result holds true when $g(x) \in T$. So, if $f(x) \in S \cap T$, then $x \in S \cap T$. Similarly, we have $g(x) \in S \cap T$ implies $x \in S \cap T$. So, by Claim 1 for $g^{-1}(x)$, we have $g(x) \in S \cap T$. $\square$ Now we shall finish the proof by noting that $f(x) \in S$ and $g(x) \in T$ implies $x \in S \cap T$ by Claim 1 which in turn implies $x \in S \cap T$ which implies $g(x) \in S \cap T$ by Claim 2 leading to a contradiction. $\blacksquare$