Let $G$ be a finite group and let $C$ be a cyclic subgroup such that $N\_G(C) = C$. Such a subgroup is in particular a Carter subgroup and Vdovin (\*) has shown that any two such subgroups are conjugate in an arbitrary finite group. Moreover, Zhang (\*\*) showed that a group with cyclic self-normalizing subgroup cannot be non-abelian simple.


I could not come up with a counterexample to the following question so far:



```
Is a finite group with self-normalizing cyclic subgroup already solvable?

```

In the paper by Zhang he mentions that the theorem he proves had been a long-standing conjecture. However, he does not give any reference. Can you point me to some previous work on this conjecture or who first formulated it?


(\*) Vdovin, E.P., [Carter subgroups of finite almost simple groups](https://doi.org/10.1007/s10469-007-0010-6). Algebra Logika 46, No. 2, 157-216 (2007), [arXiv:math/0602156](https://arxiv.org/abs/math/0602156)


(\*\*) Zhang, Guangxiang, [On self-normalizing cyclic subgroups](https://doi.org/10.1016/0021-8693(89)90250-0). J. Algebra 127, No.2, 255-258 (1989). 



Finite groups with cyclic self-normalizing subgroup
$\mathrm{P\Sigma L}(2,8)$ is a counterexample. All its cyclic groups of order $6$ are self-normalising. (They fall into two conjugacy classes.)


(This the automorphism group of $\mathrm{PSL}(2,8)$, it is $\mathrm{PSL}(2,8)$ extended by a field automorphism.)


EDIT: Examples do not seem to be plentiful. The above is the smallest one. Among the class of groups with trivial soluble radical, the next three are $\mathrm{Aut(Suz(8))}$, $\mathrm{P\Sigma L}(2,2^5)$ and $\mathrm{P\Sigma L}(2,2^7)$, with self-normalising cyclic subgroups of order $12$, $10$ and $14$, respectively. These are all the examples in the class up to order sixteen million.


