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Given 2 functions: $f: X \to X$ and $g: X \to X$,

  1. $f$ and $g$ are one-to-one
  2. $f$ and $g$ are fixed point free
  3. composite functions $f \circ g$, $\; g \circ f$, $\; f^2 \equiv f \circ f\;$ and $\;g^2 \equiv g \circ g$ are fixed point free

Is it true that, any composite functions constructed from $f$ and $g$, namely,

$ T \equiv f^{m_k} \circ g^{n_k} \circ f^{m_{k-1}} \circ g^{n_{k-1}} \dots f^{m_1} \circ g^{n_1} \circ g$

($m_i, n_i \in \mathbb{N}_0 \equiv \{0, 1, 2, \dots\}$, $\; i, k \in \mathbb{N} \equiv \{1, 2, 3, \dots\}$)

are also fixed point free?

Phil
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1 Answers1

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No.

Let $X=\{0,1,2\}$ and let $g(x) = f(x) = x + 1 \mod 3$

Yanko
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  • Oh I see. How silly I am. Thanks! – Phil Jul 24 '23 at 20:34
  • What about if I add one more condition? "There is a partition of $X$ into two subsets, and $f$ and $g$ map each of the two subsets into the other one." – Phil Jul 24 '23 at 20:44
  • @Phil So maybe write $X ={x_1,x_2,x_3,y_1,y_2,y_3}$ and let $f(x_i) = y_{i+1 \mod 3}$ and $f(y_i) = x_{i-1 \mod 3}$, and $g=f$. – Yanko Jul 25 '23 at 10:20