# Classification of groups in which the centralizer of every non-identity element is cyclic

In which classes of groups is it feasible to classify those groups in which the centralizer of every non-identity element is cyclic?

• I suspect that the word "finite" is missing. Without the finiteness assumption, this is an utterly hopeless problem, for instance, all torsion-free hyperbolic groups will belong to this class. – Misha Apr 26 '13 at 17:24
• (comment about this being hw deleted, I realized that it is a little harder than I thought). This is answered in Theorem 10 of the following paper : rose-hulman.edu/math/MSTR/MSTRpubs/1991/RHIT-MSTR-1991-05.pdf – Andy Putman Apr 26 '13 at 17:37
• Andy, for finite groups I believe that groups where all centralizers are abelian can also be classified. – i. m. soloveichik Apr 26 '13 at 17:55
• Re solovei's last comment, these are the finite CT groups (CT standing for commutative transitive). I think the relevant name here is Suzuki? – Yemon Choi Apr 26 '13 at 18:40
• @solovei : People would have reacted better if you had done two things. First, you should have phrased it as a question rather than as an imperative; as it was written, it "sounded" like homework. Second, you should have written a bit more. I'm fine with relatively terse questions, but this question needed more discussion than you gave. By the way, I voted to reopen. – Andy Putman Apr 26 '13 at 19:49

If $G$ is a finite group in which the centralizer of every nonidentity element is cyclic, then as Geoff Robinson pointed out, the Fitting subgroup $F = F(G)$ is cyclic, and $G/F$ is cyclic. One can say more, however. Suppose that $G$ is not itself cyclic, so $F < G$. Let $x$ be an element whose image modulo $F$ generates $G/F$, so $x \ne 1$. Let $C = C_G(x)$, so $C$ is cyclic, and since $x \in C$, we have $FC = G$. Since both $F$ and $C$ are cyclic, their intersection centralizes both $F$ and $C$, so is central in $G$. If this intersection contains a nontrivial element $y$, then $G =C_G(y)$, so $G$ is cyclic, contrary to assumption. Thus $F \cap C = 1$, and so $G$ can be constructed as a semidirect product of $F$ acted on by $C$.

One can say still more. Suppose some nonidentity element $c \in C$ commutes with some nonidentity element $f \in F$. Then $C_G(f)$ contains $F$ and also contains $c$. Since $c \not\in F$, we see that $C_G(f)$ strictly contains $F$. This is impossible, however, because $C_G(f)$ is cyclic, and yet $F$ is its own centralizer in $G$. This shows that the action of $C$ on $F$ is "Frobenius", and in particular, $|C|$ divides $|F|-1$ so $|F|$ and $|C|$ are coprime.

Conversely, if we start with an arbitrary finite cyclic group $F$, we can build all possible finite groups $G$ satisfying the cyclic-centralizer assumption and such that $F(G) = F$. Do this as follows. Let $d$ be the GCD of all of the numbers $p-1$ for primes $p$ dividing $|F|$, and let $C$ be any cyclic group of order dividing $d$. Then $C$ has a unique Frobenius action on $F$ and the semidirect product $G$ of $F$ by $C$ will have the desired property. The key to checking that $G$ does have this property is that every element of $G$ either lies in $F$ or is conjugate to an element of $C$, so it is enough to check that centralizers of nonidentity elements of $F$ and of $C$ are cyclic. These centralizers are respectively $F$ and $C$.

Later edit: Geoff Robinson points out that in general, the condition that all Sylow subgroups of a finite group are cyclic does not imply that $|G:F(G)|$ and $|F(G)|$ are coprime. It is interesting, however, that $|G:G'|$ and $|G'|$ must be coprime. It follows that every group with all Sylows cyclic can be constructed as a semidirect product of cyclic groups with coprime orders. Without the centralizer-cyclic condition, however, the corresponding action need not be Frobenius.

Well, in a finite group, if the centralizer of each non-identity element is cyclic, then each Sylow subgroup is cyclic ( for every prime)- just take a Sylow $p$-subgroup $P$ and a non-identity element $z \in Z(P),$ so that $P \leq C_{G}(z)$ is cyclic by hypothesis. A finite group $G$ in which each Sylow subgroup is cyclic is solvable in a very strong sense, and this needs no deep classification theorem, only elementary transfer. If $p$ is the smallest prime divisor of $|G|,$ then $N_{G}(R) = C_{G}(R)$ for each $p$-subgroup $R$ of $G,$ and then $G$ has a normal $p$-complement by Frobenius' normal $p$-complement theorem (or a transfer theorem of Burnside will do in this case). It follows by induction that a Sylow $q$-subgroup, say $Q,$ of $G$ is normal, where $q$ is the largest prime divisor of $G,$ and then $G/Q$ inherits the same property- in any case, the normal $p$-complement constructed earlier may be assumed to be solvable by induction. In fact, one can go further: since $G$ is solvable, $F = F(G)$ is cyclic, and we have $C_{G}(F) \leq F$ (the last statement is a general property of solvable groups). Hence $F$ is cyclic, and $G/F$ is Abelian, being embedded in the automorphism group of a cyclic group. Hence $G/F$ is also cyclic because of the structure of Sylow subgroups of $G$. Thus a finite group in which every centralizer of a non-identity element is cyclic. is itself metacyclic. More information about prime divisors would be needed to say more about the structure of $G.$

Later edit: Marty Isaacs' answer brings out an interesting point. In a (necessarily metacyclic) finite group $G$ whose Sylow subgroups are all cyclic, it need not be the case that $[G:F(G)]$ and $|F(G)|$ are coprime (an example of this is provided by letting a cyclic group of order $9$ act as an automorphism order $3$ on a cyclic group of order $7$. This allows the construction of a group of order $63$ with Fitting subgroup of order $21$). The extra condition of the original question, that all centralizers are cyclic, illustrates that these groups have more restricted structure than groups with all Sylow subgroups cyclic.

• As Mark Sapir and others have hinted, infinite groups with this property, and no further restrictions, seem pretty intractable. Tarski monsters have the property, for example. – Geoff Robinson Jan 25 '14 at 18:19

Gromov asked whether there is a torsion-free finitely generated group which is not word-hyperbolic and has no solvable Baumslag-Solitar subgroup $BS(1,n)$ (which do not have the property you ask for). So for finitely generated torsion-free groups, your question is as difficult as Gromov's question.

For hyperbolic groups with torsion, this question reduces to whether the elements of finite order have this property. In turn, this is equivalent to asking for the finite elements to act fixed-point free on the boundary, and for the finite subgroups (of which there are finitely many conjugacy classes) to satisfy the property (which reduces to the comment by @Andy Putman and the answer of @Geoff Robinson). So in some sense maybe your question can be answered for the class of hyperbolic groups. Maybe one could also answer your question for the class of finitely generated solvable groups.

On the other hand, for general finitely generated groups, there are the Tarski Monsters and the free Burnside groups of odd exponent. For infinitely generated groups, I imagine there could be many more possiblities.

• Doesn't $BS(2,3)$ contain no solvable Baumslag--Solitar subgroup? – HJRW Jan 25 '14 at 23:20
• @HJRW: it contains $\mathbb{Z}^2$. In fact, it can be regarded as an HNN extension of the trefoil knot group, identifying the cyclic subgroups of the 2 exceptional fibers. – Ian Agol Jan 26 '14 at 0:10
• Good point!{}{} – HJRW Jan 26 '14 at 2:14
• In fact, Rips found (torsion-free) examples of finitely generated but infinitely presented (and hence not hyperbolic) subgroups of hyperbolic groups in the early '80s. I think Gromov may have asked whether every such finitely presented group is word-hyperbolic. A (torsion-free) counterexample was found by Noel Brady. I think the correct statement is now 'For torsion-free group of type $F_3$, your question is as difficult as Gromov's question.' – HJRW Jun 26 '15 at 8:53
• PS Ian, I know you know this. But I thought it worth setting the record straight. – HJRW Jun 26 '15 at 8:54