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From group axioms, any $x$ in a group commutes with the identity, with itself, and with its inverse. By a "highly noncommutative" group is meant one for which these are the only cases of commuting elements. I found that $S_3$ (permutations of $\{1,2,3\}$) has this property, and am looking for other highly noncommutative finite groups. (Or infinite ones.) The only easy thing I noticed is that such a group should not contain any element $x$ of order greater than $3$ since if so $x$ commutes with $x^2$ and $x^2$ is neither the identity, nor $x$, nor the inverse of $x.$ However I couldn't find any such groups other than $S_3$ [though I may just not be seeing some obvious candidates].

In a comment, user moonlight has considered free products of copies of the integers mod 2 or mod 3, which also seem to me to be examples. [I didn't check all the details for these examples...]

Note: The integers mod $3$ is another example: The only distinct nonidentity elements are $1,2$ and these form an inverse pair ($1+2=0$). Also the integers mod 2 are an example, there not being distinct nonidentity elements to check anyway.

coffeemath
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    Here's a thought: call a group $G$ "quite non-commutative" iff for all $h,g \in G$, either $h$ and $g$ belong to a monogenerated subgroup of $G$, or else $h$ and $g$ fail to commute. – goblin GONE Sep 08 '15 at 08:23
  • @goblin Also an interesting definition, but includes more groups such as $\mathbb{Z}.$ The latter is not "very noncommutative" since e.g. $1$ and $2$ are distinct and commute but neither is the identity $0$ and they don't form an inverse pair. – coffeemath Sep 08 '15 at 08:33
  • I wonder what the motivation for this definition is? It seems very unnatural to me, and this is the reason you can only find a few pathological examples. If you think it is "ok" for $x$ to commute with itself, surely it would also be ok for $x$ to commute with its powers (since this is just repeated application of the former rule and also follows directly from the group axioms). – moonlight Sep 08 '15 at 10:11
  • @moonlight The fact that $x\cdot x=x \cdot x$ is immediate, just because both sides are the same computation. I agree that allowing commuting powers is "natural" in that it can be shown easily from group axioms, but was interested in only the commuting facts coming from the list of group axioms. Allowing also commuting powers seems to me to be user goblin's definition, which is an interesting one anyway but not what I had in mind. – coffeemath Sep 08 '15 at 10:15
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    I see. Goblin's definition still seems much more better motivated to me, because (i) there does not seem to be a reason to disallow repeated application of group axioms (ii) there exists no canonical list of group axioms. – moonlight Sep 08 '15 at 11:11
  • @moonlight Your point (i) is right, but a judgement call. As for point (ii) I think the axioms are almost universally closure, associativity, existence of an identity (two sided), and two sided inverse for each $x$. True, easy use of the associative law shows powers of a fixed $x$ commute, but I just thought of going with only the commutative assertions made in the initial axioms. – coffeemath Sep 08 '15 at 14:41
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    Yes, but definitions are always a judgement call. We generally strive for well-motivated and "natural" ones, because otherwise we could make a bazillian definitions and study their implications without ever finding anything useful. Another reasonable axiom systems for groups is for them to be a monoid in which all equations (in $x$) of the form $ax=b$ or $xa=b$ is solvable. – moonlight Sep 08 '15 at 14:59
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    Btw, it seems to me that your property is preserved under free products. So if you also consider infinite groups, then free products of any number of copies of $\mathbb Z_2$ and $\mathbb Z_3$ qualify. – moonlight Sep 08 '15 at 15:14
  • Let us continue this discussion in chat. – coffeemath Sep 08 '15 at 16:35

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You have indeed found all the examples.

If $|G| = 2^a3^b$ and either $a>1$ or $b>1$ then $G$ has a subgroup of order either $4$ or $9$ by Sylow's theorems. Such a subgroup is abelian and clearly contradicts the assumptions of being "very noncommutative".

Tobias Kildetoft
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  • A subgroup of order 4 or 9, if cyclic, is not "very noncommutative" since then there is an element of order more than $3$, and otherwise is $Z_2+Z_2$ or $Z_3+Z_3,$ and in these cases the choices $a=(1,0),b=(0,1)$ are distinct nonidentity elements which commute and do not form an inverse pair. [I guess this is already clear, but I wanted to be sure, so your answer indeed proves it. Note I guess one should also include the integers mod 2, in which case there aren't distinct nonidenty elements anyway, so it is trivially "very noncommutative". +1. – coffeemath Sep 08 '15 at 09:06