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The question: If $G$ is a finite group and $g \in G$ is an element of even order. Then can we colour the elements of $G$ with two colours in such a way that $x$ and $gx$ have different colours for each $x \in G$?

I wanted to verify this solution of mine.

First, $x \sim y$ if $x = g^my$ for some $ m \in \mathbb{Z}$ then this defines an equivalence relation on $G$.

  1. $x \sim x$ if we take $m = 0$.
  2. If $x \sim y$, then we have $y = g^mx$ for some $m \in \mathbb{Z}$. Then, $x = g^{-m}y$, so $y \sim x$.
  3. If $x \sim y$ and $y \sim z$, then $y = g^mx$ and $z = g^ny$ for some $m, n \in \mathbb{Z}$. That is, $z = g^{m+n}x$, so $x \sim z$.

$G$ is partitioned into $g$-orbits, so to speak. Let the order of $g$ be $2n$. In each orbit, say we colour $x$ red, $gx$ blue, $g^2x$ red, ... , $g^{2n-1}x$ blue, $g^{2n}x = x$ red we have coloured the elements of $G$ as required.

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  • $\begingroup$ This seems reasonable, but I think it might be simpler to consider the equivalence relation that $x\sim y$ if $x=g^my$ for some even number $m$. $\endgroup$
    – MJD
    Dec 10, 2017 at 16:26
  • $\begingroup$ I don't understand where we should go from there, if we do consider this equivalence relation. $\endgroup$
    – Anu
    Dec 10, 2017 at 16:38

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