The following definition has arisen naturally in two papers of mine. The papers are on rather unrelated topics; of course they are within my narrow interests, so there's some symbolic dynamics connection, but really in both cases I just needed to find good "generic elements / finite subsets" of the group, and I can't help but wonder if this has appeared elsewhere, or is even something I know but am just not recognizing.



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> Definition. A topological group $G$ is **splendid** if for all compact $C \subset G$, there exists $g \in G$ such that $gCg \cap C = \emptyset$.
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I'll mention some sufficient conditions for being splendid and give some examples. First, abelian groups are usually splendid unless obviously not splendid, and this passes to preimages, as follows:



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> Lemma. Let $G$ be a topological group that has a continuous abelian (surjective) quotient $(H,+)$ such that $\{2h \;|\; h \in H\} \leq H$ is non-compact. Then $G$ is splendid.
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For simplicity let's now concentrate on discrete infinite groups, so compact means finite. The previous lemma shows that e.g. free abelian groups, free groups and Thompson's $F$ are splendid, since they have positive rank abelianizations. Every finitely-generated infinite discrete left-orderable amenable group is indicable by a result of Morris, thus splendid by the previous lemma. (You can generalize easily to locally indicable, thus all discrete left-orderable amenable groups.)


Not every group is splendid: the infinite dihedral group $\langle a, b \;|\; a^2 = b^2 = e \rangle$ with the discrete topology is not splendid, namely pick $C = \{e, a\}$. Besides the above lemma, a way to show splendedness is that every non-splendid group that has an infinite abelian subgroup is at least a bit little dihedral (I worked this out just for this post, so take with a grain of salt):



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> Lemma. Suppose $G$ is a non-splendid discrete infinite group. If $H \leq G$ is abelian, then there is a finite-index subgroup $H' \leq H$ and some $a \in G$ such that $h^a = h^{-1}$ for all $h \in H'$.
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Proof: Let $C$ be the finite set proving non-splendidness. Then $hCh = C \neq \emptyset$ for all $h \in H$.
Since $H$ is abelian, $H$ acts on $G$ from the left by $h \* g = hgh$.
Since $h\*C \cap C \neq \emptyset$, writing $S\_a \leq H$ for the stabilizer of $a \in C$ and $T\_{a,b}$ for the transporter $\{h \in H \;|\; ha = b\}$, we have
$$ H = \bigcup\_{a, b \in C} T\_{a, b} = \bigcup\_{a, b \in C \\ T\_{a,b} \neq \emptyset} h\_{a, b} S\_a $$
for some choices $h\_{a,b} \in H$. By a result of Neumann, some $H' = S\_a$ has finite index in $H$, and for $h \in H'$ we have $hah = a \implies h^{-1} = aha^{-1}$. Square.


In particular a group is splendid if there is an element of infinite order such that no power of it is conjugate to its inverse.
For example this lemma applies to Thompson's $T$ (which has no nontrivial abelian quotients so the first lemma does not apply): Pick an element $f$ that fixes a unique interval $I \subset S^1$ and has derivative $>1$ after the right endpoint of $I$. Then $\langle f \rangle$ is of infinite order and every positive power $f^i$ has the same property as $f$, while no negative power $f^{-1}$ does. This property is preserved under conjugacy, so no $f^i$ is conjugate to its inverse.



> 
> Question. Does this have a more standard, even if possibly less splendid, existing name? Alternative characterizations that are easier to verify?
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You can refine the definition as follows, for discrete groups: say $G$ is **$k$-splendid** if the splendidness condition holds for all $|C| \leq k$. Every infinite group is $1$-splendid, while the dihedral group is not $2$-splendid. A characterization of $k$-splendidness for small $k$ would also be of interest.


A list of my favorite groups and whether or not they are splendid is also of interest to me, I tried to compile one, but after working out those Thompson's's I got stuck on the Grigorchuk group.


*Neumann, B. H.*, Groups covered by finitely many cosets, Publ. Math., Debrecen 3, 227-242 (1954). [ZBL0057.25603](https://zbmath.org/?q=an:0057.25603).


*Morris, Dave Witte*, [**Amenable groups that act on the line.**](http://dx.doi.org/10.2140/agt.2006.6.2509), Algebr. Geom. Topol. 6, 2509-2518 (2006). [ZBL1185.20042](https://zbmath.org/?q=an:1185.20042).



Splendid groups
**edit January 18th, 2021**


Final nail in the f.g. splendid coffin. Apparently, the dihedral group and its variants are indeed the only non-splendid groups. So maybe I should've called non-splendid groups splendid and vice versa, because this would roll off the tongue better then. I of course originally thought non-splendid groups would usually be some crazy monsters, and that the dihedral group was kind of an outlier.


Let's say an automorphism $\phi : H \to H$ of a group $H$ **flips** it if $\phi(h) = h^{-1}$ for all $h \in H$. Obviously a flipping automorphism can only exist on an abelian group, since $a^{-1}b^{-1} = (ba)^{-1} = \phi(ba) = \phi(b)\phi(a) = b^{-1}a^{-1}$.



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> Theorem. A finitely-generated group $G$ is non-splendid if and only if it admits an internal automorphism that flips some non-trivial finite-index normal torsion-free abelian subgroup.
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Proof. We know from the previous edit (from yesterday) that any f.g. non-splendid group $G$ is virtually abelian, and a non-trivial finite-index normal torsion-free abelian subgroup implies virtual abelianity, so we may restrict to virtually abelian $G$.


Let $H \leq G$ be any non-trivial finite-index normal torsion-free abelian subgroup (so $H \cong \mathbb{Z}^d$ for some $d \geq 1$), which exists by intersecting the conjugates of any non-trivial finite-index torsion-free abelian subgroup. We show that $H$ is flipped by an inner automorphism of $G$ if and only if $G$ is non-splendid. Because the choice of $H$ is arbitrary, this shows the a priori stronger fact $\exists H: P \implies \mbox{non-splendid} \implies \forall H: P$. Let $F \subset G$ be the finitely many coset representatives for $H$.


Suppose first that conjugation by $t \in G$ flips $H$. Write $h^t = t^{-1} h t$. Pick $C = \{ f^{-1} t, tf \;|\; f \in F \}$. Now, for any $g \in G$ we can write $g = h f$ for $h \in H, f \in F$, and then
$$ g C g \ni hf \cdot f^{-1}t \cdot hf = h t h f = t \cdot h^t h \cdot f = tf \in C. $$
So $G$ is not splendid.


Suppose then that for all $g \in G$, conjugation by $G$ does not flip $H$. Then there exists $h \in H$ which is not flipped by any $g \in G$, and in fact we can find $h$ such that conjugation by any $g \in G$ takes $h$ very far from $h^{-1}$. More precisely, for each $R$ there exists $h \in H$ such that $d(h^g, h) \geq R$ in the right-invariant word metric $d$ of $G$.


There are many ways to see the above, I guess you can geometrize and it's a basic lemma about finite reflection groups. A direct proof is also easy: Observe that it's enough to bound the word metric of $H$ since the distortion is bounded since $H$ is finite index. We only need to consider conjugation by each $f \in F$. Now take for each $f \in F$ an element $h\_f$ not flipped by $f$, and consider products of the form $h\_{f\_1}^{n\_1} \circ h\_{f\_2}^{n\_2} \circ \cdots \circ h\_{f\_{|F|}}^{n\_{|F|}}$ where the $n\_i$ have different scales. Basically, take $h\_{f\_1}$ not flipped by $f\_1$, so $h\_{f\_1}^{n\_1}$ is conjugated very far from $(h\_{f\_1}^{n\_1})^{-1}$, and take a massive $n\_1$. If also $f\_2$ does not flip $h\_{f\_1}$, we continue to $f\_3$, if $f\_2$ does flip $h\_{f\_1}$ then it does not flip $h\_{f\_1}^{n\_1} \circ h\_{f\_2}^{n\_2}$ where $n\_2$ has a smaller scale, but $f\_1$ does not flip this either since $n\_1$ is much bigger than $n\_2$. And so on. (In the geometric situation I suppose you can replace this by Baire or the fact $\mathbb{R^d}$ is not a union of finitely many $(d-1)$-dimensional subspaces.)


We can now use the elements $h$ given by the two paragraphs above to prove splendidness. Let $C$ be arbitrary, and suppose $C \subset B\_R$, the ball of radius $R$ in the right-invariant word metric of $G$ around identity. Let $h$ be such that $d(h^a, h^{-1}) > 2R$ for all $a \in G$. If $hCh \cap C \neq \emptyset$ then $h^a h = a^{-1} h a h = a^{-1} b$ for some $a, b \in C$, so $d(h^a, h^{-1}) \leq 2R$, a contradiction. So $G$ is splendid. Square.


**edit January 17th, 2021**



> 
> Theorem. Suppose $G$ is a f.g. discrete group which is not splendid. Then $G$ is virtually abelian.
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Proof. We know from the first version of my answer (see below after the text **original**) that $G$ is virtually nilpotent. Then it's well-known that $G$ has a finite-index torsion-free nilpotent subgroup $H$. Let $C$ be the finite set contradicting splendidness.


First, let's think about what splendidness says when $H$ is a torsion-free subgroup. For any $g \in H$ and $i \in \mathbb{N}$, we have $g^i a g^i = b$ for some $a, b \in C$. Now, pigeon says $g^i a g^i = b$, $g^{i+k} a g^{i+k} = b$ for some $k > 0$, and we get $g^i a g^i = g^{i+k} a g^{i+k}$, in other words $g^{-k} = a g^k a^{-1}$. Clearly this $k$ is bounded, and by taking the least common multiple of the $k$ that occur, we find $a \in C$ such that for all $g \in H$, $(g^k)^a = g^{-k}$ (true for either convention for the $g^a$ notation, because inversion is an involution).


Note that in particular $H$ is now a torsion-free nilpotent group which admits an automorphism such that for all $g \in H$, $(g^k)^a = g^{-k}$. I claim that if $H$ is not abelian, this is impossible, which will conclude the proof.


To see this, suppose $H$ is not abelian. I claim that in this case there is a three-dimensional Heisenberg group inside: We use the convention $[x,y] = xyx^{-1}y^{-1}$ if it matters. Let $(H\_i)\_i$ be the lower central series of $H$, $H\_n \neq 1, [H\_n, H] = 1$. Take $1 \neq c = [a,b] \in H\_n$ a non-trivial commutator, where $a \in H\_{n-1}, b \in H$. Now, $c$ is superlinearly distorted in $\langle a,b,c \rangle$, so certainly $a$ and $b$ cannot satisfy any identity. If $\langle a,b,c \rangle$ satisfies an identity that is not true in the Heisenberg group, then commuting elements to normal form $a^k b^\ell c^m$ we see that $c$ satisfies an identity on its own, contradicting the torsion-freeness assuption.


Now it suffices to show that the three-dimensional Heisenberg group $H = \langle a,b,c \;|\; [a,b] = c, [a,c], [b,c] \rangle$ does not admit an automorphism $\phi : H \to H$ satisfying $\phi(g^k) = g^{-k}$ for all $g \in H$. This is geometrically obvious, go around a big cycle on the $\langle a,b \rangle$-quotient, and you jump up by the area in the central direction. If you apply the inverse elementwise, you get a cycle of the same area in the same direction (just reordered). So you don't flip the direction.


In algebra, take $[a^{km}, b^{km}] = c^{k^2 m^2}$. Now apply $\phi$ to get
$$ [a^{-km}, b^{-km}] = [\phi(a^{km}), \phi(b^{km})] = \phi([a^{km}, b^{km}]) = \phi(c^{k^2 m^2}) = c^{-k^2 m^2} $$
But
$$ [a^{-km}, b^{-km}] = [a^{km}, b^{km}]^{b^{km} a^{km}} = [a^{km}, b^{km}] = c^{k^2 m^2} $$
since $[a^{km}, b^{km}]$ is central. So $c^{-k^2 m^2} = c^{k^2 m^2}$ which is the desired contradiction. Square.


Presumably you could spell out in more detail what the situations are where you get non-splendidness. I suppose it's just that some element flips the torsion-free part of the finite-index abelian subgroup. Also, the proof essentially uses the observation that non-splendidness implies essentially that some element conjugates things to their inverses, giving that inversion is almost an automorphism. This is strange since it is of course a *flip* automorphism, which sounds like it should ultimately be the reason that non-splendid groups must be virtualy abelian. Maybe you could cut out the middle man (ICC quotients) by working more on the algebra.


**original**


Joshua Frisch pointed out to me in private communication that ICC $\implies$ splendid, which seems to give a good partial answer to the question. Let me restrict to f.g. infinite discrete groups $G$.


Let $\mathcal{D}$ be the class of quotients of the fundamental group of the Klein bottle $\langle h, a \;|\; h^a = h^{-1} \rangle$, where $h$ is mapped to an element of infinite order and $a$ to a nontrivial element. For example the quotient where $a^2 = e$ is added gives the dihedral group.



> 
> Theorem. If $G$ f.g. infinite non-splendid, then $G$ is virtually nilpotent, and contains a group from $\mathcal{D}$.
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This is presumably not a characterization, but at least it reduces the problem to virtually nilpotent groups, which are much better understood than general ones. First, there's the following result of [1] (which is indeed a consequence of Neumann's theorem, but there's a nice algebraic trick that I missed).



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> Theorem. If $G$ is an f.g. ICC group, then for every finite $Z \subset G$ there exists $g$ such that all solutions to $gxg^{\pm} = y$ with $x, y \in Z$ (if they exist) satisfy $x = y = e$.
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This gives ICC $\implies$ splendid. There's a minor complication with involutions, so I give a proof:



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> Theorem. If $G$ is an f.g. ICC group, then it is splendid.
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Proof. Suppose ICC, and fix some symmetric generating set. Let $n$ be arbitrary and take $Z = B\_N$ for $N = 2n$. By the previous theorem, for some $g \in G$, among $x, y \in Z$ the equation $gxg^{\pm} = y$ can only have solutions with $x = y = e$. If $g$ is not an involution, there aren't any such solutions either, and we are done.


If $g$ is an involution, then take any $h$ of word norm exactly $n$, and consider $gh$ instead. Now if $gh \cdot k \cdot gh = k'$ for $k, k' \in B\_n$, then $g \cdot hk \cdot g = k' h^{-1}$ and here $hk, k'h^{-1} \in Z$, and this cannot happen by the assumption on $g$. So we have $gB\_ng \cap B\_n = \emptyset$. Square.


Next, there's a theorem of [1,2] (see also [3]).



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> Theorem. Every f.g. infinite group is either virtually nilpotent, or has a non-trivial ICC quotient.
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And also let's prove a lemma.



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> Lemma. If $G$ has a splendid quotient, it is splendid.
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Proof. Let $\phi : G \to H$ be the quotient map, and $C \subset G$ be finite. Then $\phi(C) = C'$ is also finite, and thus $hC'h \cap C' = \emptyset$ for some $h \in H$. Letting $\phi(g) = h$, we have $\phi(gCg) \cap \phi(C) = hC'h \cap C' = \emptyset$, so $gCg \cap C = \emptyset$ as well, and this means $G$ is splendid. Square.


We obtain the following:



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> Theorem. Every non-splendid group is virtually nilpotent.
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Proof. If $G$ is not virtually nilpotent, then it has a non-trivial ICC quotient. That quotient is then splendid, and thus $G$ is splendid. Square.


Now we can prove the theorem I stated at the beginning of the post.


Proof. An infinite f.g. nilpotent group contains $H \cong \mathbb{Z}$, which is abelian. By the lemma in my post, there exists $a \in G$ and $h \in H \setminus \{e\}$ such that $h^a = h^{-1}$ for all $h \in H$. Then $h$ and $a$ give the quotient, taking $n$ the order of $a$. Square.


[1] *Erschler, A.; Kaimanovitch, V. A.*, Arboreal structures on groups and the associated boundaries, Arxiv [https://arxiv.org/abs/1903.02095]](https://arxiv.org/abs/1903.02095%5D)


[2] *Duguid, A. M.; McLain, D. H.*, (FC)-nilpotent and (FC)-soluble groups, Proc. Camb. Philos. Soc. 52, 391-398 (1956). [ZBL0071.02204](https://zbmath.org/?q=an:0071.02204).


[3] *McLain, D. H.*, [**Remarks on the upper central series of a group**](http://dx.doi.org/10.1017/S2040618500033414), Proc. Glasg. Math. Assoc. 3, 38-44 (1956). [ZBL0072.25702](https://zbmath.org/?q=an:0072.25702).


[4] *Frisch, J.; Ferdowsi, P. V.*, Non-virtually nilpotent groups have infinite conjugacy class quotients, Arxiv [https://arxiv.org/abs/1803.05064]](https://arxiv.org/abs/1803.05064%5D)


