Let $G$ be a finite simple group and $\tau_G = \{ o(x) : x \in G\}$. Does there exist $d_1, d_2 \in \tau_G$ that satisfy the following:

  • $d_1 < d_2$ and $d_1$ does not divide $d_2;$
  • for $x, y \in G$ with $o(x) = d_1$ and $o(y) = d_2$, we have $xy = yx$?

I have tried by taking an alternative group $A_5$ and $A_6$ but was unable to reach any conclusion. I would be thankful for any kind of help.

  • $\begingroup$ As far as I understand i don’t think (1) holds for any group, but I could be wrong. $\endgroup$ Jun 25 at 14:57
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    $\begingroup$ It does, @blakedylanmusic; consider $$\Bbb Z_2\times\Bbb Z_3.$$ It has a subgroup of order two and a subgroup of order three. $\endgroup$
    – Shaun
    Jun 25 at 14:59
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    $\begingroup$ @Shaun true… after I made the comment I was starting to realize it wasn’t correct. Another example would be $S_6$, the symmetric group on 6 letters, where there exists a permutation of order 5 and another permutation of order 3. $\endgroup$ Jun 25 at 15:03

1 Answer 1


The second property does not hold in any simple group for any $d_1,d_2 \in \tau_G \setminus \{1\}$.

If is did, then $x \in G$ with $o(x)=d_1$ would commute with all elements in $G$ of order $d_2$. But the elements of order $d_2$ generate a normal subgroup of $G$, which by simplicity would be all of $G$, so $x \in Z(G)$, contradiction.

  • $\begingroup$ Thanks everyone and especially Derek, I was trying to get this one but was stuck there. $\endgroup$
    – Struggler
    Jun 25 at 15:15

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