5
$\begingroup$

For a finite group $G$ one defines the spectrum $\omega(G)$ as the set of element orders of $G$. The set $\omega(G)$ is uniquely determined by the subset $\mu(G)$ consisting of those elements of $\omega(G)$ that are maximal with respect to the divisibility relation.

I would like to know whether for a finite non-abelian simple group $G$ one has $|\mu(G)| \geq 3$.

If this is the case, is there a proof that works for all finite simple groups of Lie type (without treating each family differently)?

$\endgroup$
  • $\begingroup$ Did you consider $A_n$, $n\ge 5$? $\endgroup$ – Mark Sapir Aug 7 '17 at 19:27
  • $\begingroup$ I did a quick computation with GAP and $\mu(A_n)$ seems to grow with $n$. It seems only $A_5$ and $A_6$ achieve the lower bound with $\mu(A_5) = \{ 2, 3, 5 \}$ and $\mu(A_6) = \{ 3,4, 5\}$. I also remember having checked this for the sporadic groups. $\endgroup$ – scalar Aug 7 '17 at 19:39
  • $\begingroup$ $\mu(A_n)$ should be easy to compute by hand. $\endgroup$ – Mark Sapir Aug 7 '17 at 19:46
  • 2
    $\begingroup$ It's not clear to me that $\mu(A_n)$ is easy to compute, but showing $\mu(A_n)\ge 3$ is easy. If there are 3 distinct primes in $[n/2,n]$, then $|\mu(A_n)|$ since it contains 3 elements divisible by these elements, which have to be distinct. It seems to hold if $n=13,14$ and $n\ge 17$. It still works if we have three odd prime powers $\le n$ such that the sum of any two of these is $>n$. This holds for $n=7$ and $n\ge 9$. The remaining cases are ok since $2,3,5\in\mu(A_5)$, $3,4,5\in\mu(A_6)$ and $4,5,7\in\mu(A_8)$. $\endgroup$ – YCor Aug 7 '17 at 20:42

Your Answer

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

Browse other questions tagged or ask your own question.