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For a finite generating set $A$, use the symmetric alphabet $\Sigma =A\cup A^{-1}$. The Cayley graph has a vertex for each $g\in G$ and an oriented edge labelled $a$ from $g$ to $ga$. Give every edge length one. Its vertex word metric isTwo finitely presented groups are quasi-isometric when their Cayley graphs, with these path metrics, are quasi-isometric. The choice of finite generating set is immaterial, as the following proof shows; the relators identify vertices but do not otherwise enter the metric definition.

Given finite generating sets $A,B$ of the same group, put $L=\max (1,{\max }_{a\in A^{\pm 1}}|a{|}_B)$ and $M=\max (1,{\max }_{b\in B^{\pm 1}}|b{|}_A)$. Replace each letter in a shortest $A$-word by a fixed $B$-word to obtain $|g{|}_B\le L|g{|}_A$. Reversing the roles gives $|g{|}_A\le M|g{|}_B$. Therefore the vertex identity map satisfiesIt is onto, and hence a quasi-isometry. Map each edge to the fixed word path for its label, or pass through the uniformly dense vertex sets, to obtain the corresponding graph quasi-isometry with a bounded additive error. This proves equivalence of finite word metrics and independence of the two finite group presentations.

The Milnor–Švarc lemma applies to an isometric properly discontinuous group action that is also a cocompact group action on a nonempty proper metric space that is a geodesic metric space. It concludes that the group is finitely generated and, for a basepoint $x$ and any finite generating set of a group $S$, the orbit mapHere $d_S$ is the word metric, and a quasi-isometry means a map satisfying two-sided coarse distance bounds with a uniformly coarsely dense image. A free action is not required: finite stabilizers are compatible with proper discontinuity. The properness and cocompactness hypotheses ensure that only finitely many group elements move a chosen basepoint by the bounded amount used to generate the group. Replacing a finite generating set by another does not change the conclusion, by equivalence of finite word metrics.

The two commutation relations make $c$ central, and the remaining relation is $ab=cba$. Repeatedly move $a$ past $b$, keeping the central powers in front. Induction first on $s$ and then on $r$ givesThereforeThis is the commutator identity in the integer Heisenberg group. The order of the factors fixes the positive sign of the exponent.

To check the claimed infinite order, map the generators to matricesTheir commutator is $I+E_{13}$, which commutes with both other matrices, so all defining relations hold. The image of $c^m$ is $I+m{E}_{13}$; hence $c$ has infinite order. This realizes the presentation in the Integer Heisenberg group and is sufficient to certify the cyclic subgroup, without needing to prove faithfulness of the whole representation.

With the ambient word metric from $\{a,b,c{\}}^{\pm 1}$, the displayed commutator givesFor any finite generating set $T$ of $⟨c⟩$, write its elements as powers of $c$ and set $B=\max \{|j|:c^j\in T\}$, with inverses included. Since $T$ generates an infinite cyclic group, $B\ge 1$. A word of length $m$ in $T$ has exponent at most $Bm$ in absolute value, soA quasi-isometric embedding of the inclusion would require $|c^{k^2}{|}_H\ge {\lambda }^{-1}|c^{k^2}{|}_T-\epsilon$. Combining these bounds would give $4k\ge k^2/(B\lambda )-\epsilon$ for every $k$, which is impossible. Thus the inclusion of the central infinite cyclic subgroup is not a quasi-isometric embedding. Changing the ambient finite generating set only changes multiplicative constants, by equivalence of finite word metrics. This is the quadratic distortion of the Heisenberg center.

The word length is the cost of building a group element from the allowed generators of a group. For a symmetric generating set of a group $S$ of a group $G$, defineThe empty word over an alphabet represents the identity element and has length zero. A representative with this least number of letters labels a shortest graph path from $1$ to $g$ in the Cayley graph. Thus word length minimizes over all representatives, rather than counting the letters in whichever expression we happened to write down.^[ref]

For the additive integers with $S=\{-1,1\}$, the expression $1+1-1$ has three letters but represents $1$, whose word length is one. With $T=\{-2,-1,1,2\}$, the integer $5$ has word length three, realized by $2+2+1$. The Cayley graphs of the integers visualize why these are minimal. A freely reduced word is always shortest in a free group with its free generators and their inverses; free reduction alone need not give a shortest representative in a general group.

Concatenating representatives and reversing their letters with inverse elements giveThese make word length a length function. A finite $S$ gives finite word metric balls, since only finitely many words over an alphabet have length at most a fixed radius. The definition itself also works for infinite symmetric generating sets of a group; the bounded-metric example in equivalence of finite word metrics shows what finiteness adds.