Let $q=p^f$ be a power of the prime $p$ and let $\alpha$ be an inner diagonal automorphism of ${^2}{\operatorname{E}_6(q^2)}$, i.e., an element of $({^2}\operatorname{E}_6)_{\mathrm{ad}}(q^2)$ in the notation of [1, pp. 39ff.].

Question: Is $|\{x\in{^2}{\operatorname{E}_6(q^2)}\mid\alpha(x)=x\}|=|\operatorname{C}_{{^2}{\operatorname{E}_6(q^2)}}(\alpha)|\geq q^D$ for some explicit, absolute constant $D>0$?

Some remarks on this Question:

  1. When replacing ${^2}{\operatorname{E}_6(q^2)}$ by any other finite simple group of Lie type ${^t}{X_r(q^t)}$, the answer to the above Question is "yes" by known results:
    • For simple Chevalley groups and simple classical groups, this follows from [2, Theorem 6.15] (the value of the universal constant $A$ mentioned there can be worked out from the bounds in the single cases proved in [2, Section 6]).
    • All finite simple groups of Lie type ${^t}{X_r(q^t)}$ that are neither Chevalley groups nor classical groups nor ${^2}{\operatorname{E}_6(q^2)}$ do not have (nontrivial) diagonal automorphisms, and so for these, the problem is about centralizers of inner automorphisms $\alpha$ only. When $\alpha$ is semisimple (i.e., corresponds to a $p'$-element of ${^t}{X_r(q^t)}$), then $\operatorname{C}_{{^t}{X_r(q^t)}}(\alpha)$ contains a maximal torus of ${^t}{X_r(q^t)}$, which by [3, Lemma 3.3] has size at least $\frac{1}{|\operatorname{Out}({^t}{X_r(q^t)})|}\cdot(q-1)^r\geq q^{1/2}$ except possibly for some small $q$ (up to an explicitly determinable upper bound), which can be ignored. On the other hand, when $p\mid\operatorname{ord}(\alpha)$, then [4, proof of Theorem A1, p. 320] gives $|\operatorname{C}_{{^t}{X_r(q^t)}}(\alpha)|\geq q$.
  2. The "torus argument" for semisimple elements of ${^t}{X_r(q^t)}$ above also works for semisimple inner diagonal automorphisms of ${^2}{\operatorname{E}_6(q^2)}$, so one can assume that $p\mid\operatorname{ord}(\alpha)$ in the above Question.
  3. Possibly, the argument given in [4, proof of Theorem A1, p. 320] to show that centralizers of non semisimple elements of ${^t}{X_r(q^t)}$ contain some root subgroup over some extension field of $\mathbb{F}_q$ (and thus have order at least $q$) can be extended to non semisimple elements of $({^2}\operatorname{E}_6)_{\mathrm{ad}}(q^2)$, but this is not obvious to me.


[1] R.W. Carter, Finite groups of Lie type. Conjugacy classes and complex characters, John Wiley & Sons (Chichester), 1985.

[2] J. Fulman and R. Guralnick, Bounds on the number and sizes of conjugacy classes in finite Chevalley groups with applications to derangements, Trans. Amer. Math. Soc. 364(6):3023-3070 (2012).

[3] B. Hartley, A general Brauer-Fowler theorem and centralizers in locally finite groups, Pacific J. Math. 152(1):101-117 (1992).

[4] B. Hartley and M. Kuzucuoğlu, Centralizers of elements in locally finite simple groups, Proc. London Math. Soc. (3) 62(2):301-324 (1991).


1 Answer 1


The answer is `yes'. The subgroup of $C_{{^2}E_6(q^2)}(s)$ generated by all its $p$-elements is a central product of groups of Lie type over extension fields of $\mathbb{F}_q$, if $s$ is a semisimple inner diagonal automorphism. (This can be seen by writing $C_{{^2}E_6(q^2)}(s)=C_{C_G(s)}(\sigma_q\tau)$ where $G$ is the adjoint algebraic group $E_6$ over the algebraic closure of $\mathbb{F}_q$, $\sigma_q$ is the $q$th power map, and $\tau$ is a graph automorphism of $G$ of order $2$. See 4.2.2 in Gorenstein-Lyons-Solomon, The classification of the finite simple groups, Number 3, Amer. Math. Soc. Surveys and Monographs 40. $\quad$ So in the troublesome case when $\alpha$ is not semisimple, $C_{{^2}E_6(q^2)}(\alpha)$ does contain an isomorphic copy of the additive group of $\mathbb{F}_q$, as you surmise.


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