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. Algebra Logika 46, No. 2, 157-216 (2007), arXiv:math/0602156

(**) Zhang, Guangxiang, On self-normalizing cyclic subgroups. J. Algebra 127, No.2, 255-258 (1989).

  • 2
    $\begingroup$ I think $\mathrm{P\Sigma L}(2,8)$ is a counterexample. It has a self-normalising cyclic subgroup of order 6. $\endgroup$
    – verret
    Jun 26, 2017 at 10:15

1 Answer 1


$\mathrm{P\Sigma L}(2,8)$ 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.


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