# Groups where no elements commute except for the trivial cases

Let $$G$$ be a group. Write $$e$$ for its neutral element and write $$\langle g\rangle$$ for the subgroup generated by an element $$g \in G$$. Assume that $$G$$ has the following properties:

1. For all $$g\in G\setminus\{e\}$$ and $$h\in G\setminus \langle g \rangle$$ we have $$gh \neq hg$$.

2. Property 1. is non-vacuous (as it would be e.g. for $$G=\{e\}$$).

Do such groups exist? If so, do they have any interesting/important properties? Note that this is a follow-up to this very similar question in response to one of the comments there.

• Just to point out: the apparent motivation for your question suggests also considering a weaker constraint, that $gh \ne hg$ unless either $h \in \langle g \rangle$ or $g \in \langle h \rangle$. In general we can have one of these without the other, since the order of $g$ may strictly divide that of $h$ (or vice versa), and the assumption that this does not occur is what forces $G$ to have squarefree order. Thus, though the change in definition is slight, the range of solutions should be considerably wider. – Robin Saunders Jul 5 at 9:45
• @RobinSaunders, if you wish to make this into a question, I will share my reflections on it. – Andreas Caranti Jul 8 at 11:00

Let us consider a finite group $$G$$ with this property.

Let $$P \ne 1$$ be a Sylow $$p$$-subgroup of $$G$$. If $$g$$ is an element of order $$p$$ in $$Z(P)$$, then every element of $$P$$ commutes with $$g$$, so that $$P = \langle g \rangle$$.

Thus all Sylow subgroup have prime order, that is, the order of $$G$$ is squarefree.

It follows that $$G$$ is metacyclic, and actually the semidirect product of two cyclic groups (I am thinking Schur-Zassenhaus or Hall's theorems, but it might be simpler than that), which by an argument similar to the one above have to be of prime order.

It follows that the finite groups with this property are the non-trivial semidirect products of a cyclic group of prime order $$p$$ by a cyclic group of prime order $$q \mid p - 1$$.

PS This related discussion may be of interest.

To provide contrast to Andreas Caranti's answer, a Tarski monster $$p$$-group is an example of an infinite group with these properties.

If $$G$$ is a Tarski monster $$p$$-group then (by definition) every proper nontrivial subgroup is cyclic of order $$p$$. The centralizer of any nontrivial element $$g$$ is a proper subgroup (since $$G$$ has trivial center), and thus must be $$\langle g\rangle$$.

Since you haven't limited your question to finite groups, another example is the free group on $$n$$ generators, where $$n \gt 1$$.

Edited to add: As noted below in the comments, this isn't correct because $$g=x^2, h=x$$ is a counterexample (where $$x$$ is a generator of the group).

• But if $a$ is a generator, then $a\not\in\langle a^2\rangle$, while $a$ and $a^2$ commute. – Gabe Conant Jul 4 at 14:47
• To extend on @GabeConant's comment: All elements of OP's $G$ must have finite prime order, because if the order of $a$ is infinite or a proper multiple of prime $p$, then $g:=a^p$ and $h:=a$ are a counterexample to the property. – Magma Jul 4 at 14:59
• That's true. I was focused on the elements. I'd delete the answer but I think the associated comments are valuable and I don't want to lose them. – Robert Shore Jul 4 at 15:08
• Still, free groups are an interesting example of groups enjoying a property that is probably more natural, namely that the centralisers of non-trivial elements are cyclic, see the very interesting discussion here. – Andreas Caranti Jul 5 at 9:32

The group $$S_3$$ is what you're after. Besides, since its order is $$6$$, it's easy to verify that those conditions hold.

• Thanks! But is it the only one? Or at least the only finite one? – Mars Plastic Jul 4 at 14:33
• I didn't check it, but my guess is that every $S_n$ (with $n>2$) will do. – José Carlos Santos Jul 4 at 14:34
• $S_n$ doesn't work for $n > 3$, since the transpositions $(12)$ and $(34)$ commute. On the other hand, the dihedral group $D_{2p}$ works for any prime $p$. – Magma Jul 4 at 14:47