2
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ As far as I understand i don’t think (1) holds for any group, but I could be wrong. $\endgroup$
    – blakedylanmusic
    Jun 25 at 14:57
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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. $\endgroup$
    – blakedylanmusic
    Jun 25 at 15:03

1 Answer 1

Reset to default
8
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Thanks everyone and especially Derek, I was trying to get this one but was stuck there. $\endgroup$
    – Struggler
    Jun 25 at 15:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.