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Let $G$ be a finite group. Consider $X = \{[x,y] \mid (\lvert x \rvert, \lvert y \rvert) = 1, x,y \in G \}$ and $K = \left< X \right>$. Suppose that $\vert a^X \vert = m$, for all $a \in G$. If $y \in K$, exists $n \in \mathbb{Z}$, $n>0$ such that $y^n \in Z(G)$? If yes, this $n$ depends only on $m$?

Notation: $[x,y]$ - denotes the commutator of the elements $x$ and $y$; $(\lvert x \rvert, \lvert y \rvert)$ - denote the greatest common divisor of the elements $x$ and $y$; $Z(G)$ - denotes the centre of the group $G$.

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What is $\vert a^X \vert$? –  Robert Lewis Aug 13 '13 at 20:00
    
By the way, you should use \vert instead of \mid for absolute values, otherwise the spacing is awkward. –  M Turgeon Aug 13 '13 at 20:05
    
@RobertLewis I would assume that $|a^X|$ is the size of the set of elements of $G$ that are fixed by the elements of the $X$. –  rckrd Aug 13 '13 at 20:14
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"Fixed by elements of $\,X\,$ "? Fixed under what ? I'd rather say that $\,a^X\,$ is the set of all the conjugates of $\,a\,$ by elements of $\,X\,$, but why should we guess? Let the OP address these concerns and clarify his own question...or not. –  DonAntonio Aug 13 '13 at 23:34
    
What makes you think that this would be true? –  Alexander Gruber Aug 14 '13 at 13:06

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