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Let $G$ be a finite group such that for all $x\neq e$,

$$C_G(x)=\langle x\rangle$$

Is there any classification of such groups ?

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Suzuki's work handles the case where $\langle x \rangle$ are replaced by abelian subgroups. Higman's work handle's the case where we only require this for different primes ($x$ cannot centralize $y$ if $x$ and $y$ have coprime order). –  Jack Schmidt May 21 at 20:21

2 Answers 2

up vote 18 down vote accepted

Yes, these are the cyclic groups of order $p$ and the non-abelian groups of order $pq$.

Reduction 1: These groups have no elements of prime squared order, lest elements of prime order have larger centralizers.

Reduction 2: The Sylow p-subgroups are cyclic, lest (for p odd) they have an abelian subgroup of order $p^2$, or (for p=2) the have a central involution that is not the entire Sylow.

Combining these two results, we get the sylows are all prime order, so we get a so-called group of square-free order.

Such groups were classified by Zassenhaus, and a nice description is in Hall's Theory of Groups. They are all metacyclic. However, this group has no cyclic subgroups of composite order, so those two cyclic groups both have prime order.

Hence the group has order $pq$. Obviously the abelian one doesn't work.


Cleaner version: Let $G$ be such a group. If $A \leq G$ is abelian, then $A \leq C_G(x)$ for every $x \in A$. Hence $A$ must be the cyclic group generated by all of its non-identity elements. Let $P$ be a Sylow $p$-subgroup, and let $z \in Z(P)$. Since $P \leq C_G(z)$, we must have $P=\langle z \rangle$ is abelian, so $P$ too is generated by all of its non-identity elements. Hence every Sylow $p$-subgroup of $G$ has order at most $p$, so $|G|$ is a square-free number. By Zassenhaus (Hall's Theory of Groups Theorem 9.4.3 on page 146), $G$ has cyclic subgroups $H,K$ with $K=[G,G]$ such that $G=HK$. Again $H,K$ must be generated by their non-identity elements, so (if not the identity) must have distinct prime orders. This gives that either $G=H$ is cyclic of prime order, or $G=H\ltimes [G,G]$ is a non-abelian group of order $pq$ for distinct primes $p<q$. One verifies that all such groups do in fact satisfy the $C_G(x)=\langle x \rangle$ conclusion.


Generalizations: If one only requires that elements of order $p$ not centralize elements of order $q$, then one gets the groups with “discrete prime graph” (prime graphs are studied by our user Alexander Gruber) and classified in Higman (1957). If one instead requires that elements in certain abelian subgroups only are centralized by those abelian subgroups, then one gets the CA groups classified in Suzuki (1957) and Brauer–Suzuki–Wall (1958).

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Thanks :-) I will check Suzuki's work. –  mesel May 21 at 20:29

Same basic idea as in the answer by Jack.

First of all, it is immediate that every nontrivial element of $G$ must have prime order. Furthermore, $p^2$ does not divide the order of $G$ for any prime $p$, as otherwise an element of order $p$ would be contained in a subgroup of order $p^2$ (which is abelian).

Thus $G$ has squarefree order. It is a basic fact about groups of squarefree order that $G$ has a normal $p$-Sylow $P$, where $p$ is the largest prime dividing $|G|$ (you could prove this with Burnside's normal complement theorem). Then $\operatorname{Aut}(P)$ is abelian, so $N_G(P)/C_G(P) = G/P$ is abelian. Since $|G|$ is squarefree, $G/P$ is cyclic. Because every nontrivial element of $G$ has prime order, this is true for $G/P$ as well. Hence either $G/P$ is trivial or has prime order.

It follows then that either $G$ has prime order, or is nonabelian of order $pq$ for distinct primes $p$ and $q$.

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@JackSchmidt: $G/P$ is cyclic. If its order is divisible by two distinct primes $p$ and $q$, then it contains an element of order $pq$, which gives an element of composite order in $G$. –  Mikko Korhonen May 22 at 6:50
    
So silly, thanks! :-) –  Jack Schmidt May 22 at 15:38

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