Let $G$ be a finitely generated group with finite generator $S$, and define the metric $ d_S : G \times G \to \mathbb{Z} \cap [0, \infty)$ by
$$d_S(x, y) = \min \left\{ \ell \geq 0 : \exists s_1, \ldots, s_\ell \in \left( S \cup S^{-1} \right) \setminus \left\{ e \right\} \; \left( x^{-1} y = s_1 \cdots s_\ell \right) \right\} ,$$
i.e. $d_S$ is the word metric associated to $S$. The metric $d$ will automatically be left-invariant, i.e. $d_S(x, y) = d_S(gx, gy)$ for all $g, x, y \in G$, since $(g x)^{-1} (g y) = x^{-1} g^{-1} g y = x^{-1} y$.
However, this metric need not always be right-invariant. For an example of the word metric not being right-invariant, consider the example of $G = F_2$, i.e. the free group on two generators, with (free) generator $S = \{a, b\}$. Then $d_S(e, b) = 1$, but $d_S(e a, b a) = d_S \left( e, a^{-1} b a \right) = 3$, so the "canonical" word metric on the free group with two generators is not right-invariant.
On the other hand, if $G$ is abelian, then every word metric is right-invariant. Right-invariance is equivalent to the conjugation action $g \cdot x = g^{-1} x g$ of $G$ on itself satisfying $d_S(e, g \cdot x) = d_S(e, x)$ for all $g, x \in G$, since $d_S (xg , y g) = d\left(e, g^{-1} y g \right)$. So conjugation needs to preserve word length, and of course the conjugation action for an abelian group is trivial. Alternatively, if $G$ is finite, and $S = G$, then $d_S$ is the discrete metric (i.e. $d_S(x, y) = \delta(x, y)$), and thus is also right-invariant. This trick works for any discrete group if we drop our assumption that $S$ be finite, but I'd rather consider the finite generator case.
My question: is there a straightforward characterization of when the word metric is both left- and right-invariant? My guess would be that every word metric is right-invariant if and only if $G$ is abelian, but I can't prove this either. This seems like the kind of problem that should be elementary to solve, but alas, I haven't solved it.
Alternatively, is there some sense in which the word metric will be "almost right-invariant", where we can place bounds on the extent to which the word metric fails to be right-invariant?
Thanks in advance for your help!
EDIT 1: Here's a partial answer based on @Mariano Suárez-Álvarez's suggestion, at the end of which are some related follow-up questions.
Claim: Let $S$ be a finite generator for $G$, and set $\tilde{S} = \left(S \cup S^{-1} \right) \setminus \{e\}$. Let $g \star x = g^{-1} x g$ be the action of $G$ on itself by conjugation (note: this is a right action which I foolishly wrote on the left; thanks @Captain Lama for pointing this out). Then $d_S$ is right-invariant if and only if $s \star t \in \tilde{S}$ for all $s, t \in \tilde{S}$.
Proof: The necessity is obvious, since if there existed $s, t \in \tilde{S}$ for which $s \star t \not \in \tilde{S}$, then we'd have $d_S(s, t s) = d_S \left( e, s \star t \right) \neq 1 = d_S (e, t)$. Now we claim it's sufficient.
First, we remark that if $\tilde{S}$ is closed under conjugation by itself, then $\tilde{S}$ is closed under conjugation by all elements of $G$, i.e. $g \cdot s \in \tilde{S}$ for all $s \in \tilde{S}, g \in G$. This follows by writing an arbitrary element $g$ of $G$ as a product of elements of $\tilde{S}$ and applying an inductive argument. Moreover, we remark that if $\tilde{S}$ is closed under conjugation, then $|x \star y| \leq |y|$ for all $x, y \in G$, where $|\cdot|$ denotes word length. Let $x = s_1 \cdots s_\ell, y = t_1 \cdots t_m$, where $s_i, t_j \in \tilde{S}$ and $|y| = m$. Then
\begin{align*}
|x \star y| & = |x \star (t_1 \cdots t_m)| \\
& = |(x \star t_1) \cdots (x \star t_m)| \\
& = \left| t_1' \right| \cdots \left| t_m' \right| \\
& = m & = |y| ,
\end{align*}
where $t_j' = x \star t_j \in \tilde{S}$.
Thus conjugation cannot increase word length. We claim, however, that conjugation preserves word length, i.e. that $|x \star y| = |y|$ for all $x, y \in G$. Assume for contradiction that there exist $x, y \in G$ such that $|x \star y| < |y|$. Let $y' = x \star y$. Then $\left| x^{-1} \star y' \right| = |y| > \left|y'\right|$. This contradicts our earlier claim that $\left|z \star w\right| \leq |w|$ for all $z, w \in G$.
Thus if $\tilde{S}$ is closed under conjugation, then conjugation preserves word length. From here, it follows that if $x, y, g \in G$, then
$$d_S(x g, y g) = \left| (x g)^{-1} (yg) \right| = \left| g^{-1} x^{-1} y g \right| = \left| g \star \left( x^{-1} y \right) \right| = \left| x^{-1} y \right| = d_S(x, y) .$$
This concludes the proof.
This criterion encompasses some non-abelian groups. For example, as we mentioned, every finite group can be realized in this way by taking $S = G \setminus \{e\}$.
This proof can be elaborate on to show the word metric is "almost right-invariant" in the sense of "Lie Groups and Error Analysis" by Schiff and Shnider: Set
$$M = \max_{s, t \in \tilde{S}} |s \star t| \in \mathbb{N} .$$ For each $g \in G$, set $\rho(g) = M^{|g|}$. Consider $x, y \in G$, and let $g = s_1 \cdots s_\ell, x^{-1} y = t_1 \cdots t_m$ for $s_i t_j \in \tilde{S}$, where $\ell = |g|, m = \left| x^{-1} y \right| = d_S(x, y)$. Then
\begin{align*}
d_S(x g, y g) & = \left| g \star \left( x^{-1} y \right) \right| \\
& = \left| (g \star t_1) \cdots (g \star t_m) \right| \\
& \leq |g \star t_1| + \cdots + |g \star t_m| \\
& = \sum_{j = 1}^m |(s_1 \cdots s_\ell) \cdot t_j| .
\end{align*}
Applying an inductive argument, we can conclude that $|(s_1 \cdots s_\ell) \cdot t_j| \leq M^\ell$ for all $j$, so
\begin{align*}
d_S(x g, y g) & \leq \sum_{j = 1}^m |(s_1 \cdots s_\ell) \cdot t_j| \\
& \leq \sum_{j = 1}^m M^\ell \\
& = M^\ell \cdot m \\
& = M^{|g|} \left| x^{-1} y \right| & = M^{|g|} d_S(x, y) .
\end{align*}
Edit 2: As @Sean Eberhard pointed out, this estimate can be cleaned up significantly to get
\begin{align*}
d_S(x g, y g) & = \left| g^{-1} \left( x^{-1} y \right) g \right| \\
& \leq \left| g^{-1} \right| + d_S(x, y) + |g| \\
& = 2|g| + d_S(x, y) \\
& \leq (2|g| + 1) d_S(x, y) ,
\end{align*}
so in general $d_S(x g, y g) \leq (2|g| + 1) d_S(x, y)$. However, this is estimate is in general sharp. Consider the case where $G$ is a free group of rank $\geq 2$ generated freely by $S$. If $g \in G$, then we can write $g = s_1 \cdots s_\ell$ for $s_1, \ldots, s_\ell \in S \cup S^{-1}$, where the word is in reduced form so $|g| = \ell$. Let $x$ be an element of $S \cup S^{-1}$ such that $x \not \in \left\{ s_1, s_1^{-1} \right\}$. Then $g^{-1} x g = s_\ell^{-1} \cdots s_1^{-1} x s_1 \cdots s_\ell$, and this is the reduced form, so $\left| g^{-1} x g \right| = 2|g| + 1$. Thus
$$d_S(g, xg) = 2|g| + 1 = (2|g| + 1) |x| = (2|g| + 1) d_S(e, x) .$$
So if $G$ is a nonabelian free group and $S$ is a free generator for $G$, then for every $g \in G$ exists $x \in G$ such that $d_S(g, xg) = (2|g| + 1) d(e, x)$, meaning that $2|g| + 1$ is the optimal Lipschitz constant for $x \mapsto g^{-1} x g$.
Assuming all the above is correct, I think there are at least two questions that are reasonable to ask:
- What groups admit right-invariant word metrics, or equivalently, which groups admit finite generators closed under conjugation? Clearly every finitely generated abelian group fits this bill, but I've yet to cook up an infinite non-abeliean example
- For which groups is every word metric right-invariant? Clearly every finitely generated abelian group fits this bill, but I've yet to cook up a non-abelian example, including the finite case.
Best Answer
As Mariano answered in the comments, the metric is right-invariant if and only if $S$ is conjugation-invariant.
Answers to your follow-up questions:
Suppose $S$ is a finite symmetric conjugation-invariant generating set for $G$. The conjugation action of $G$ on $S$ induces a homomorphism $G \to \mathrm{Sym}(S)$ whose kernel has index at most $|S|!$. But the kernel is $Z(G)$. Therefore $G$ has a finite-index (and finitely generated) center.
Conversely, suppose $Z(G)$ has finite index in $G$. Then all conjugacy classes are finite (they have order at most the index of $Z(G)$). Let $S$ be a finite symmetric generating set for $G$. Then $S' = \bigcup_{g \in G} S^g$ is a finite symmetric conjugation-invariant generating set.
There are plenty of uninteresting examples like $G \times \mathbf{Z}^d$ for any finite group $G$. A nonsplit example is given by the nontrivial semidirect product $C_3 \rtimes \mathbf{Z}$ (which is nonsplit as an extension of its center). It is possible to classify all examples in terms of data consisting of a finite group $G$ and some outer automorphisms of $G$.
One nonabelian example is the quaternion group $Q_8 = \{\pm1, \pm i, \pm j, \pm k\}$, whose conjugacy classes are $1, -1, \{\pm i\}, \{\pm j\}, \{ \pm k\}$. Therefore every set of the form $\{x, x^{-1}\}$ is conjugation-invariant, so every symmetric generating set is conjugation-invariant.
Conversely suppose $G$ has a conjugacy class $C$ not of the form $\{x\}$ or $\{x, x^{-1}\}$. Let $S$ be a finite symmetric generating set for $G$. If $S$ is not conjugation-invariant, we are done. Otherwise $S \cap C$ is either $\emptyset$ or $C$. If $S \cap C = \emptyset$ then pick $x \in C$ and consider instead $S' = S \cup \{x, x^{-1}\}$: this finite symmetric generating set contains only part of $C$ so it is not conjugation-invariant. Suppose on the other hand that $C \subseteq S$. Let $x \in C$. Since $S$ generates $G$ there must be some $s \in S$ such that $x^s \notin \{x, x^{-1}\}$. Note moreover that $s \ne x^{\pm1}$. Let $S' = S \setminus \{x, x^{-1}\}$. Then $s, x^s \in S'$, so $x = (x^s)^{s^{-1}} \in \langle S'\rangle$, so $S'$ still generates $G$. Again it contains only part of $C$, so it is not conjugation-invariant.
Therefore we want to classify finitely generated groups such that all conjugacy classes have the form $\{x\}$ or $\{x, x^{-1}\}$. Note that any subgroup of such a group is automatically normal: such a group is called a Dedekind group, and they are known to be either abelian or of the form $Q_8 \times B \times D$ where $B$ is an elementary abelian $2$-group and $D$ is a torsion abelian group with all elements of odd order. But if $x \in D$ is nontrivial then $\{(\pm i,1,x)\}$ is a nontrivial conjugacy class of $Q_8 \times B \times D$ not of the required form, so $D$ must be trivial. Specializing to the finitely generated case, it follows that $G$ is either abelian or of the form $Q_8 \times B$ where $B$ is a finite elementary abelian $2$-group.