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.
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.