7
$\begingroup$

Can a finite simple group $G$ have an element $x$ of order 8 such that $x^2$ is conjugate to $x^{-2}$ but $x$ is not conjugate to any element of $\{x^3,x^5,x^7\}$?

In other words, does this fusion pattern force a normal subgroup of index 2?

Edit: Noam Elkies gave an example with no normal subgroup of index 2. In his example, the Sylow 2-subgroups have non-trivial proper strongly closed subgroups (direct factors). Is there a non-simple example that at least has no strongly closed subgroups?


I believe any counterexample $G$ has a Sylow 2-subgroup of order at least 128, and $G$ has order at least $10^{10}$. There might be a simple group out there that just works, but I don't think it is in the ATLAS's character tables, and I don't think it is very small.

I believe simple transfer arguments are doomed to fail (examples show $x \in P \cap [G,G]$ can occur with this fusion pattern, even though I've always had $P \cap [G,G] < P$), but I would love to see a technique similar to transfer that works.

Such an argument would be difficult since in Noam Elkies's example, $P$ is its own hyperfocal subgroup and $P$ has no fusion-normal subgroups (though $P_1 \times 1$ and $1 \times P_2$ are strongly closed of course). Must such a $G$ always have a Sylow 2-subgroup with a non-identity proper strongly closed subgroup?


Here are the related patterns:

There are plenty of examples of $G$ that do have normal subgroups, but for every other fusion pattern in a cyclic subgroup of order 8, I've either found a simple group with that pattern or proved that no group (simple or not) has such a pattern.

The other patterns that can occur:

  • $\newcommand{\PSL}{\operatorname{PSL}}\PSL(2,17)$ has $x \sim x^{-1}$ (and so $x^2 \sim x^{-2}$) but $x^3 \not\sim x \not\sim x^5$.
  • $\PSL(3,3)$ has $x \sim x^3$ (and so $x^2 \sim (x^2)^3=x^{-2}$) but $x^{-1} \not\sim x \not\sim x^5$.
  • ${}^2F_4(2)'$ has $x \sim x^5$ and (independently) $x^2 \sim x^{-2}$, but $x^{-1} \not\sim x \not\sim x^3$
  • $\PSL(3,5)$ has $x \sim x^5$ but $x^3 \not\sim x \not\sim x^{-1}$ and $x^2 \not\sim x^{-2}$
  • $M_{12}$ has $x \sim x^{-1} \sim x^3 \sim x^5$ and $x^2 \sim x^{-2}$

The others that are impossible:

  • If $x \sim x^3 \sim x^5$ then we also have $x \sim (x^3) \sim (x^3)^5 = x^{-1}$, so that this "two out of three" fusion is not possible (neither are the other "2 out of 3" fusions)

I am happy with any ideas. If you can show $G$ cannot be simple, that is enough. If you can find a $G$ that has no normal subgroups of index 2, that is enough. (Actually it is still enough for me, I now have a new aspect to work on.) If you know of any papers that have techniques that might work, that would also be wonderful.

I've previously asked on math.se.

$\endgroup$
3
  • 1
    $\begingroup$ If you have two simple groups $G_1,G_2$ with elements $x_1,x_2$ of order $8$ with $x_i^2 \sim x_i^{-2}$, and $x_1$ conjugate only to $x_1^3$ while $x_2$ is conjugate only with $x_2^5$, then $(x_1,x_2) \in G_1 \times G_2$ gives an example. Admittedly $G_1 \times G_2$ is not simple; still, it does lack index-$2$ subgroups. $\endgroup$ Feb 4, 2014 at 17:21
  • $\begingroup$ Thanks! This means that transfer is doomed to fail, as well as most things I know as "fusion arguments", since that group is fusion simple. $\endgroup$ Feb 4, 2014 at 17:32
  • $\begingroup$ I've adjusted the question so that there is something left to answer. For me personally your answer is great: it pointed out the futility of a wide variety of arguments, leaving me only one argument left, which I can now concentrate on. Of course, if there is such a simple group then I'd love to hear about that :-) $\endgroup$ Feb 4, 2014 at 19:13

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.