# Fusion pattern in a cyclic subgroup of order 8

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.

• 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. – Noam D. Elkies Feb 4 '14 at 17:21
• 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. – Jack Schmidt Feb 4 '14 at 17:32
• 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 :-) – Jack Schmidt Feb 4 '14 at 19:13