Given a finite group $G$, let $\eta(G)$ denote the number of distinct sizes of orbits on $G$ under the action of ${\rm Aut}(G)$.

It happens that there are infinitely many non-abelian finite simple groups $G$ such that $\eta(G) = 5$. For example, this holds for the groups ${\rm PSL}(2,p)$ for prime numbers $p \geq 7$.

In contrast, is it true that $G = {\rm A}_5$ is the only non-abelian finite simple group such that $\eta(G) = 4$? -- GAP computations suggest this may be true.

Remark: I know that among the alternating groups, the sporadic groups and the simple groups of Lie type except for those of type $^2{\rm A}$, ${\rm A}$, ${\rm B}$, ${\rm C}$, ${\rm D}$, $^2{\rm D}$ and $^3{\rm D}_4$, there are no further groups $G$ with $\eta(G) \leq 4$.

Update (April 30, 2020): Meanwhile this question has 3 deleted answers.

  • $\begingroup$ I think this is an open problem , but i didn't remeber when i accrossed that $\endgroup$ – user147204 Apr 28 '20 at 22:19
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    $\begingroup$ I guess for a group of Lie type of rank $r$ over $\mathbb{F}_q$, if you take $r$ large enough and $q$ large enough, then $\mu(G)>4$ -- there will be regular semisimple elements with centralizers that are different maximal tori... And a bit of argument with Zsigmondy primes will ensure that the outer automorphisms don't yield any "accidental" coincidences in orbit size. Doing this more carefully, and lobbing some unipotent elements into the mix, might get you down to something like $r\leq 6$ and $q\leq 11$, say. Just a guess... $\endgroup$ – Nick Gill May 1 '20 at 14:41
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    $\begingroup$ By the way your remark suggests you know this to be true for the unitary groups. Is this correct, or have you omitted them in error? $\endgroup$ – Nick Gill May 1 '20 at 14:42
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    $\begingroup$ @ Nick Gill Indeed what one can see from some properties of such groups, is that if we take r large enough and q large enough, then μ(G)>4.... As to the Unitary groups, no I haven't dealt with these groups... Thank you for reminding me..I edited the question. $\endgroup$ – Leyli Jafari May 2 '20 at 6:16
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    $\begingroup$ I would have thought you have a reasonable chance of pushing this approach to get a full proof (that the answer is YES). For instance, I looked at Kleidman's paper on ${^3\!D_4}(q)$ where, amongst other things, he lists the maximal tori. There are seven classes up to isomorphism, so there should be no problem in excluding this case. There is also a handy paper by Buturlakin and Grechkoseeva that describes the cyclic structure of maximal tori for the classical groups. As soon as you are past rank 3, you have lots and lots of these. You'd need to be careful with small $q$ but still... $\endgroup$ – Nick Gill May 2 '20 at 9:47

Let me see if I can write down a proof that the answer is YES for most of the $A_n$-groups. The method should work for the rest of the $A_n$-groups and, indeed, all of the other groups you mention. Specifically, I'll prove

Proposition: If $G={\rm PSL}_n(q)$ with $n\geq 5$ and $q\geq 2$, then $\mu(G)>4$.

Proof: In what follows, I'll write $q=p^f$ where $p$ is a prime, $f$ a positive integer. Then $|{\rm Aut}(G)|/ |{\rm PGL}_n(q)|=2f$. Recall that a primitive prime divisor of $p^{df}-1$ is a prime that divides $p^{df}-1$ but not $p^k-1$ for any integer $k<df$. Zsigmondy's theorem asserts that such a prime always exists unless $df=2$ or $(p,df)=(2,6)$.

An easy argument shows that if $r$ is a primitive prime divisor of $p^{df}-1$, then $r>df$. It's written down as Lemma 2.7 of a paper of mine with Azad and Britnell. This is important because it means that primitive prime divisors of $p^{df}-1$ do not divide $2f$ provided $df>1$.

So now our job is to find five elements, $g_1,\dots, g_5\in G$, with different orbit sizes under ${\rm Aut}(G)$. Write $o(g_i)$ for the the orbit size of $g_i$. In what follows we note down what makes each $o(g_i)$ definitely different to the others.

  • Let $g_1=1$. Then $o(g_1)=1$.
  • Let $g_2$ be central in a Sylow $p$-subgroup of $G$. Then $o(g_i)$ is not divisible by $p$.
  • Let $g_3$ be an element whose centralizer is a maximal torus in ${\rm PGL}_n(q)$ of size $\frac{q^n-1}{q-1}$. Then $o(g_3)$ is not divisible by a ppd of $q^n-1$.
  • Let $g_4$ be an element whose centralizer is a maximal torus in ${\rm PGL}_n(q)$ of size $q^{n-1}-1$. Then $o(g_4)$ is not divisible by a ppd of $q^{n-1}-1$.
  • Let $g_5$ be an element whose centralizer is a maximal torus in ${\rm PGL}_n(q)$ isomorphic to $(q^{n-2}-1)\times(q-1)$. Then $o(g_5)$ is not divisible by a ppd of $q^{n-2}-1$.

Some remarks:

  1. the existence of tori of this size can be seen directly, but is written down explicitly in the paper of Buturlakin and Grechkoseeva.
  2. The fact that a ppd of $q^{n-2}-1$ does not divide $q^n-1$ follows from the fact that $n\geq 5$ and the fact that $gcd(q^{n-2}-1, q^n-1)$ divides $q^2-1$. Similarly for the other pairs of ppds.
  3. We are implicitly using the lower bound on ppd's used above -- so that the action of $|{\rm Out}(G)|$ doesn't mess things up.
  4. In theory one should check what happens when $(q,n)\in\{(2,6), (2,7), (2,8)\}$ -- here Zsigmondy's theorem fails for one of the tori mentioned above and a ppd does not exist. But I've excluded $q=2$ so this doesn't arise (see next comment).
  5. Finally the fact that there really are elements that are centralized by these maximal tori is a fairly straightforward eigenvalue argument. The only problem arises when $q=2$ and we have the torus $(q^{n-2}-1)\times (q-1)$, hence I've excluded $q=2$ from the statement of the theorem. QED

Extra remarks

  • Dealing with $q=2$ should be easy. For instance if $n$ is odd, then you can substitute that torus of size $(q^{n-2}-1)\times (q-1)$ with one of size $(q^{n-2}-1)\times (q+1)$ (maybe this will work for $n$ even too but I haven't checked).
  • Likewise $n=2,3,4$ can be done by hand (I guess).
  • The whole argument carries over to the unitary groups just by changing a few signs in the size of the tori. The other classical groups will require that you choose tori of different sizes.
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    $\begingroup$ Thanks a lot for your interesting partial answer. $\endgroup$ – Leyli Jafari May 3 '20 at 10:54

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