Must a finite group $G$ with an element $x$ of order 8 such that $x^2$ is conjugate to $x^{-2}$ but $x$ is not conjugate to any element of $\{x^3,x^5,x^7\}$ also have a normal subgroup of index 2?
I believe any counterexample $G$ has a Sylow 2-subgroup of order at least 128, and any simple counterexample $G$ has order at least $10^{10}$.
There are plenty of examples of $G$ that do have normal subgroups of index 2, but for every other fusion pattern, I've either found a simple group with that pattern or proved that no group (simple or not) has such a pattern.
The most obvious way for $x^2$ to be conjugate to $x^{-2}$ is if $x$ is conjugate to $x^{-1}$. Indeed if $x^g = x^{-1}$, then $(x^2)^g = (x^g)^2 = (x^{-1})^2 = x^{-2}$ and $x^2$ is inverted by the same conjugation that inverts $x$. Of course if $x^g = x^3$, then $(x^2)^g = (x^g)^2 = (x^3)^2 = x^6 = x^{-2}$ works as well. 5 is somewhat irrelevant: if $x^g = x^5$ then $(x^2)^g = (x^g)^2 = (x^5)^2 = x^{10} = x^2$.
The simple groups ${}^2F_4(2)'$ and ${}^2F_4(8)$ both have $x$ conjugate to $x^5$ and $x^2$ conjugate to $x^{-2}$ (where the conjugating elements of course differ for $x$ and $x^2$) without having $x$ conjugate to $x^3$ or $x^7$.
The other patterns are more common:
- $\newcommand{\PSL}{\operatorname{PSL}}\PSL(2,17)$ 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 $x^{-1} \not\sim x \not\sim x^5$.
- $\PSL(3,5)$ 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 $x^2 \sim x^{-2}$
Or impossible:
- If $x \sim x^3 \sim x^5$ then $x \sim (x^3) \sim (x^3)^5 = x^7$ and similarly for other "two of three" possibilities.
Edit: 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. If you know of any papers that have techniques that might work, that would also be wonderful.