# Existence of a finite simple group that satisfies particular properties.

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.

• As far as I understand i don’t think (1) holds for any group, but I could be wrong. Jun 25 at 14:57
• It does, @blakedylanmusic; consider $$\Bbb Z_2\times\Bbb Z_3.$$ It has a subgroup of order two and a subgroup of order three. Jun 25 at 14:59
• @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. Jun 25 at 15:03

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.