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In this MO question it was mentioned that the following fact seems to be true:

If $G$ is doubly transitive on $X$ and the one-point stabilizer $G_x$ has a non-trivial center, then $G$ is of affine type, that is, the socle is elementary abelian.

Does anyone know if it is true?

I checked it with GAP for all 2-transitive groups up to degree 2499 and I no counterexamples appeared. Here is the code for groups with maximun degree of transitivity equal to two:

for gr in AllPrimitiveGroups(NrMovedPoints, [1..2499], Transitivity, 2) do
  stab := Stabilizer(gr, 1);
  if not IsTrivial(Center(stab)) then
    soc := Socle(gr);
    if not IsAbelian(soc) then 
      Print("Counterexample of degree ", NrMovedPoints(gr), "\n");
    fi;
  fi;
od;

I am mainly interested in the following question:

Is it true that if $G$ is a 2-transitive group with simple socle then $Z(G_x)=1$?

I need only to consider groups with maximum degree of transitivity equal to two. Further, many of the groups appearing in the classification of 2-transitive groups with simple socle are easy to handle. However, I cannot prove the fact in the following cases:

  1. Socle $PSL(d,q)$, degree $\frac{q^d-1}{q-1}$, and $d\geq3$ (two actions).
  2. Socle $PSU(3,q)$, degree $q^3+1$ and $q\geq3$.
  3. Socle $Sz(q)$, degree $q^2+1$ and $q=2^{2d+1}>2$.
  4. Socle $Ree(q)$, degree $q^3+1$ and $q=3^{2d+1}>3$.

Is it true that if $G$ is a 2-transitive almost simple group with simple socle isomorphic to $PSL(d,q)$, $PSU(3,q)$, $Sz(q)$ or $Ree(q)$, then $Z(G_x)=1$?

References:

  • Cameron, Peter J. Finite permutation groups and finite simple groups. Bull. London Math. Soc. 13 (1981), no. 1, 1--22. MR0599634 (83m:20008). The table with the classification of 2-transitive groups with simple socle appears in page 157, see google books)
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  • $\begingroup$ I guess you mean $Z(G_x) = 1$ in your two questions, rather than $Z(G_x) \neq 1$... $\endgroup$ – Tom De Medts Oct 17 '14 at 15:32
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    $\begingroup$ @Leandro: If I understand the set-up correctly, the group $G_x$ here is a "Borel subgroup" (the group $B$ in the usual $(B,N)$-pair structure). So in your specified rank 1 Lie type cases corresponding to the 2-transitive assumption the center of $B$ is certainly trivial. Here $G$ is acting 2-transitively on the set of Borels, each of which is its own normalizer. $\endgroup$ – Jim Humphreys Oct 17 '14 at 20:28
  • $\begingroup$ @Jim: Thanks! Do you have a reference for this? In my case, $G=Aut(N)$ is doubly transitive on $X$ and it is the group of automorphisms of the simple group $N$, where $N$ is one of $PSL(d,q)$, $PSU(3,q)$, $Sz(q)$ or $Ree(q)$. $\endgroup$ – Leandro Vendramin Dec 3 '14 at 16:12
  • $\begingroup$ @Leandro: No, I was just calling attention to the fact that being 2-transitive is equivalent to having a rank 1 BN-pair. This might not be directly helpful in your examples but is worth thinking about if you want a general framework. (However, the literature on BN-pairs has little to say about rank 1.) $\endgroup$ – Jim Humphreys Dec 3 '14 at 17:49
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This question is connected to an extreme case of the odd analogue of Glauberman´s $Z^{\ast}$-theorem. This theorem asserts that if a finite group $G$ has no non-identity normal subgroup of order coprime to the prime $p,$ and $u$ is an element of order $p$ of $G$ which commutes with none of its other $G$-conjugates, then $u \in Z(G).$ This theorem was proved (by Glauberman) without CFSG for $p =2,$ but as far as I know all proofs to date for odd $p$ use CFSG.

Now suppose that $G$ is a doubly transitive permutation group with $F(G) =1,$ and that the point stabilizer $G_{x}$ has a central element $u$ of prime order $p$ (which will certainly happen if $Z(G_{x}) \neq 1).$ Then $G_{x} = C_{G}(u)$ and the permutation action is equivalent to that of $G$ acting by conjugation on the conjugates of $u.$

Now $C_{G}(u)$ permutes the conjugates of $u$ which commute with $u.$ By the double transitivity of the permutation action, we see that either all conjugates of $u$ commute with $u,$ or else no conjugate of $u$ other than $u$ itself commutes with $u.$ In the former case, $u \in F(G),$ which is excluded by hypothesis. Hence $u$ commutes with none of its other conjugates. By the general $Z^{\ast}$-theorem, we either have $u \in Z(G)$ or else $O_{p^{\prime}}(G) \neq 1.$ The former case is excluded as $F(G) = 1.$

Suppose then that $O_{p^{\prime}}(G) \neq 1.$ Then the image of $u$ is central in $G/O_{p^{\prime}}(G)$ and a Frattini argument yields $G = O_{p^{\prime}}(G)C_{G}(u).$

Since $u \not \in Z(G),$ there is a prime $q \neq p$ such that $u$ normalizes, but does not centralize, a Sylow $q$-subgroup, $Q$ say, of $O_{p^{\prime}}(G)$. Hence there is a $Q$-conjugate $v$ of $u$ such that $\langle u,v \rangle$ is a $\{p,q\}$-group. By the transitivity of $C_{G}(u)$ on the other conjugates of $u,$ it follows that $\langle u,w \rangle$ is a $\{p,q\}$-group for every conjugate $w$ of $u.$ It then follows that $[O_{p^{\prime}}(G),u] \lhd G$ is a non-trivial $q$-group, and that $F(G) \neq 1,$ contrary to hypothesis.

Later edit: I noticed that this also follows from Theorem D of :Guralnick, Robert M.; Robinson, Geoffrey R. On extensions of the Baer-Suzuki theorem. Israel J. Math. 82 (1993), no. 1-3, 281–297. One needs to use an earlier theorem of E. Shult. An outline proof is as follows:

We have a doubly transitive finite group $G$ with non-Abelian simple socle. Suppose that $H$ is a point stabilizer, and that $x$ is an element of prime order $p$ in $Z(H).$ Then $G$ acts doubly transitively by conjugation on the conjugates of $x$, so $C_{G}(x)$ is transitive on the remaining conjugates of $x$. There can be no other conjugate $y$ of $x$ which commutes with $x$, since if there were, all conjugates of $x$ would commute with $x$, contrary to the simplicity of the socle of $G$. By a theorem of E. Shult, there must be a $p^{\prime}$-subgroup $T$ of $G$ which is normalized, but not centralized. Hence for some $t \in T, x^{-1}x^{t}$ is a non-identity $p$-regular element. Since $C_{G}(x)$ is transitive on the remaining conjugates of $x$, we see that $x^{-1}x^{g}$ is $p$-regular for all $g \in G$. By Theorem D of the Guralnick-Robinson paper, and the argument above ( Theorem D gives $G = O_{p'}(G)C_{G}(x)$, and the argument above gives $[G,x] = [O_{p'}(G),x]$ is a $q$-group for some prime $q$, contrary to the fact that $G$ has non-Abelian simple socle).

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  • $\begingroup$ Thanks! Would you have a reference for this generalization of the Glauberman Z*-theorem? $\endgroup$ – Leandro Vendramin Feb 2 '15 at 17:25
  • $\begingroup$ Possibly the first place it was recorded in the literature was in a paper of O.D. Artemovich in 1988 (see MR0952132) $\endgroup$ – Geoff Robinson Feb 2 '15 at 17:34
  • $\begingroup$ @GeoffRobinson By F(G) (in the first answer) you mean Z(G), or, for some reason, the Fitting subgroup or what? $\endgroup$ – Giuliano Bianco Feb 11 '15 at 11:18
  • $\begingroup$ @GeoffRobinson Also in the second answer, even after the last edit, you invoke (line 3) the simplicity of G, which is not in the hypothesis. By the way it is a really illuminating answer! Thank you prof. Robinson! $\endgroup$ – Giuliano Bianco Feb 11 '15 at 11:21
  • $\begingroup$ @Giuliano Bianco : F(G) is the Fitting subgroup of G. Since G has non-Abelian simple socle, the Fitting subgroup of G must be trivial, as must Z(G). I have fixed the reference to the simplicity of G- since G has no-Abelian simpe socle, G has trivial Fitting subgroup, so non non-identity element of G commutes with all its conjugates. – Geoff Robinson $\endgroup$ – Geoff Robinson Feb 11 '15 at 14:51

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