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Any two word metrics from finite generating sets on the same group are bilipschitz equivalent. Indeed, if $S,S^{\prime }$ are finite, let $L={\max }_{s\in S}|s{|}_{S^{\prime }}$; then $|g{|}_{S^{\prime }}\le L|g{|}_S$, and the reverse inequality follows symmetrically. Apply this once to the two finite generating sets of $G$ and once to those of $H$. Composing these bilipschitz identity maps with the inclusion changes only the multiplicative and additive constants in the quasi-isometric embedding inequalities. Thus being a quasi-isometrically embedded subgroup is independent of $S$ and $T$.

Take $G={\mathbb{Z}}^2$ with the standard generating set $S=\{(1,0),(0,1)\}$, and letIntrinsic distance in $H$ between $0$ and $(n,n)$ is $|n|$, while its ambient word metric distance is $2|n|$, so $H$ is quasi-isometrically embedded. However, the ambient geodesic from $(0,0)$ to $(n,n)$ that first travels to $(n,0)$ and then to $(n,n)$ contains $(n,0)$. Its distance from the diagonal subgroup $H$ is $n$. No uniform $K$ can contain every such geodesic in the $K$-neighborhood of $H$, so $H$ is not a quasiconvex subgroup.

A finitely generated group of isometries acting properly discontinuously on a complete connected $n$-dimensional Riemannian manifold with nonnegative Ricci curvature has polynomial growth of a group of degree at most $n$. Choose $\rho >0$ so distinct points of an orbit are separated by more than $2\rho$. Let $L$ be the largest displacement of the base point by a generator. The radius-$m$ word metric ball maps into the metric ball of radius $Lm$, with fibers of size the finite stabilizer $|{\Gamma }_p|$. Disjoint radius-$\rho$ balls around these orbit points have equal volume and fit inside the ball of radius $Lm+\rho$. The Bishop-Gromov inequality bounds their number by $(1+Lm/\rho {)}^n$, giving the displayed estimate.

For a word metric from a finite generating set, the stable word length isSubadditivity and the Fekete lemma give the limit. Conjugating $g^n$ changes its length by at most a constant, so stable word length is conjugacy invariant. It satisfies ${\tau }_S(g^k)=|k|{\tau }_S(g)$. Finite-order elements have zero stable word length; in a hyperbolic group, every infinite-order element has positive stable word length.

The word metric measures how many allowed moves are needed to change one group element into another. For a finite symmetric generating set of a group $S$ with $1\notin S$, it is the graph distance between vertices of the Cayley graph, equivalentlyHere word length counts letters in a shortest representative of $g^{-1}h$. The path metric on the entire Cayley graph also permits points inside edges; the word metric is its restriction to the group itself.^[ref]

The word metric is left invariant:This corresponds to the label-preserving group action on the Cayley graph. Symmetry follows by reversing a shortest graph path; concatenating routes gives the triangle inequality.^[ref]

Right invariance can fail in a non-abelian group, because right multiplication replaces $g^{-1}h$ by $a^{-1}{g}^{-1}ha$ through conjugation. In the free group $F(x,y)$ with $S=\{x,x^{-1},y,y^{-1}\}$, we have $d_S(1,y)=1$, whereas $d_S(x,yx)=|x^{-1}yx{|}_S=3$: the latter freely reduced word has three letters. In an abelian group, conjugation fixes each element, so right invariance does hold. Cayley graphs of the integers and word metric balls in the integer lattice give examples where the word metric can be calculated explicitly.

The closed word metric ball of radius $r\ge 0$ centred at a group element $g$ isThis is a closed ball for the word metric. A finite generating set of a group gives a finite word metric ball, because at most ${\sum }_{k=0}^{\lfloor r\rfloor }|S{|}^k$ words over an alphabet can represent its elements. Different words over an alphabet may represent the same element, so this is an upper bound. Word metric balls in the integer lattice give exact counts for two choices of moves.

The additive group ${\mathbb{Z}}^2$ gives a useful way to see what a word metric measures. Let $e_1=(1,0)$ and $e_2=(0,1)$, and choose the four-move generating set of a group $S=\{\pm e_1,\pm e_2\}$. Its Cayley graph is the square lattice. Each move changes exactly one coordinate by one, so the word length is the L1 norm, and the word metric is Manhattan distance:At least this many moves are necessary, and moving horizontally then vertically attains the bound. Bridson gives this square lattice as a basic Cayley graph example.^[ref]

Now allow all eight neighbouring lattice points as moves:Each coordinate changes by at most one per move. Use diagonal moves until the smaller displacement is exhausted, then use axis moves: the resulting word length is the maximum norm, and the word metric is Chebyshev distance. From $(0,0)$ to $(2,3)$, an $S$-route has five steps, while the $T$-route $(0,0),(1,1),(2,2),(2,3)$ has three.

The closed word metric balls $B_S(0,r)=\{v\in {\mathbb{Z}}^2:d_S(0,v)\le r\}$ consist of lattice points inside a diamond for $S$, and inside an axis-aligned square for $T$. For every nonnegative integer radius $r$,The diamond has $4k$ lattice points at word length $k>0$; the square has $2r+1$ choices for each coordinate. At radius three the counts are $25$ and $49$. These are discrete closed balls for Manhattan distance and Chebyshev distance; an Euclidean distance bound would instead select lattice points inside a Euclidean disk.