This question is partly motivated by a situation which arises in modular representation theory. A finite group $G$ is said to be almost simple if $G$ has a unique minimal normal subgroup which is a non-Abelian simple group. Such a group $G$ has $S \leq G \leq {\rm Aut}(S)$ for some non-Abelian finite simple group $S$. The question then is whether such an almost simple group $G$ can contain an element $x$ of order $3$ which normalizes no non-identity $2$-subgroup of $G$? Note that if there is such an such an $x,$ then $x$ can lie in no subgroup of $G$ which is cyclic of order $6$, isomorphic to $A_{4}$, or isomorphic to $S_{4}.$

Later edit: Two simple groups which might be considered "near misses" are ${\rm PSL}(2,9) \cong A_{6}$ and ${\rm PSL}(2,27)$. For example, ${\rm PSL}(2,27)$ has only one maximal subgroup of order divisible by $6,$ which is isomorphic to $A_{4}.$ If we look directly at $A_{6}$, the fact that there is an unusual copy of $A_{5}$ containing the element $(123)(456)$ produces a non-trivial $2$-group normalized by $(123)(456)$ ( though the fact that all elements of order $3$ in ${\rm PSL}(2,9)$ are conjugate within ${\rm PGL}(2,9)$ gives a more satisfying explanation. The same argument covers ${\rm PSL}(2,3^{n})$ for all $n>1$ from the point of view of this question).

Perhaps I should remark that it is easy to produce finite groups $H$ containing an element $y$ of order $3$ normalizing no non-identity $2$-subgroup of $H$: just take any finite group $X$ containing an element $z$ of order $3$, and let $U$ be a $2$-subgroup of $X$ which is maximal subject to being normalized by $z$. If $U = 1,$ we are done. If not, then $y = zU$ is an element of order $3$ in $H = N_{X}(U)/U$ which normalizes no non-identity $2$-group of $H$.

Later edit: Bob Guralnick has pointed out to me that ${\rm SL}(2,2^{2n+1})$ is a family of simple groups where there is such an element of order 3, which changes the nature of the question. Nevertheless, such examples seem to be rather rare, and it might be informative to have a complete list of examples. Even later: Ree groups are another example, as pointed out by John Thompson.

Later yet: A complete list of such almost simple groups has now been determined by Spencer Gerhardt (a student of Bob Guralnick), in the context of answering a more general question. The results will appear in a forthcoming paper by Gerhardt.

  • $\begingroup$ I think the mentioning of ${\rm S}_4$ is redundant here. $\endgroup$ – Stefan Kohl Jan 22 '15 at 16:52
  • $\begingroup$ @StefanKohl: Yes, in the sense that if it's inside an $S_{4}$, it's inside an $A_{4}$ ( or from the point of view that if it normalizes a $2$-group, it normalizes an elementary Abelian $2$-group and breaking that into irreducible prices yields a cyclic group of order $6$ or an $A_{4}).$ But from character theory point of view, it might help to note that (if such an $x$ exists at all) there is no involution $t \in G$ such that $tx$ has order $3,4$ ( or $5$ for that matter). $\endgroup$ – Geoff Robinson Jan 22 '15 at 17:00
  • $\begingroup$ Should that be "Such a group $G$ has $S\leq G\leq \mathrm{Aut}(S)$" in line $3$? $\endgroup$ – Arturo Magidin Jan 22 '15 at 19:34
  • $\begingroup$ @ArturoMagidin :Yes, indeed. Thanks- a silly typo. I will correct. $\endgroup$ – Geoff Robinson Jan 22 '15 at 19:51
  • 2
    $\begingroup$ @JimHumphreys: Yes, I imagine the Classification will probably be needed, though maybe not its full force- however, some powerful theorems would seem to be necessary- if there were one, a simple group of odd order divisible by $3$ would have such an element, for example! $\endgroup$ – Geoff Robinson Jan 22 '15 at 22:17

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