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A Cayley graph turns multiplication in a group into travel: the chosen generators of a group are the allowed one-step moves, and a word over an alphabet becomes a route. This lets us study a finitely generated group using graph distances, closed balls, and geodesics, rather than only equations between products.

Let $G$ be a group with a finite generating set of a group $S$ satisfying $S=S^{-1}$ and $1\notin S$. Such a symmetric subset of a group includes the inverse element of every allowed move. The Cayley graph ${\mathrm{Cay}}_S(G)$ has vertex set $G$ and joins $g$ to $gs$ for each $s\in S$. We choose right multiplication for these moves. Reversing the move uses $s^{-1}$, so for graph distance we regard $g\to gs$ and $gs\to g$ as the two orientations of one undirected edge. An involution therefore gives one edge, not two parallel edges. Each edge has length one, regardless of its length in a drawing.^[ref]

A sequence $s_1,\dots ,s_n$ labels the walkIt ends at $h$ exactly when $g^{-1}h=s_1\cdots s_n$. Thus the Cayley graph is a connected graph precisely because $S$ generates $G$, and its graph distance on vertices is the word metric. Left multiplication by $a$ sends the edge from $g$ to $gs$ to the edge from $ag$ to $(ag)s$, preserving its generator label: this is a group action by graph automorphisms and isometries. Viewing groups as symmetries of geometric objects is the motivation of Clay and Margalit's introduction.^[ref]

The useful geometric object includes its allowed moves, not just its drawing. Cayley graphs of the integers show how changing the moves adds shortcuts; word metric balls in the integer lattice show how the same group can acquire visibly different closed balls. Meier's book develops this passage from group operations to graphs as its central motivation.^[ref]

For the additive infinite cyclic group $\mathbb{Z}$, the generating set of a group $S=\{-1,1\}$ produces the infinite line: each integer $n$ is adjacent to $n-1$ and $n+1$. Adding the moves $\pm 2$ gives $T=\{-2,-1,1,2\}$ and additional edges skipping one vertex. Each skip still costs one step in the word metric.

For integers $a,b$, these word metrics areThe absolute value counts unit moves, and the ceiling function counts pairs plus a possible leftover unit. Every $T$-move changes the integer by at most two, while moves of size two followed by at most one move of size one attain this bound. For example, the graph path $0,1,2,3,4,5$ has five steps for $S$, while $0,2,4,5$ has three for $T$. This calculation gives the concrete comparison

The picture shows only a finite window of each infinite Cayley graph. By contrast, $U=\{-2,2\}$ alone fails to be a generating set of a group for $\mathbb{Z}$: it generates the subgroup $2\mathbb{Z}$. Its move graph has two connected components of a graph, the even integers and the odd integers. Within either component the graph distance is $|a-b|/2$; between different components no finite graph path exists. Adding a shortcut is different from deleting the moves needed for connectivity.

A cobounded group action by isometries on a nonempty geodesic metric space admits an orbit map that is a quasi-isometry for a possibly infinite generating set of a group. If $Gx$ is $R$-dense, take $S=\{g\ne 1:d(x,gx)\le 2R+1\}$. Subdivide a metric geodesic into segments of length at most one and approximate subdivision vertices by orbit points, keeping the endpoints exact. The resulting group increments generate and giveThe first estimate includes zero displacement: a nontrivial stabilizer element has length one. No finiteness assertion about $S$ survives without properness.

For a proper subgroup $H$ of any group $G$, its complement is a generating set of a group. Choose $x\notin H$. For every $h\in H$, also $xh\notin H$, and $h=x^{-1}(xh)$ lies in the subgroup generated by the complement. That subgroup consequently contains both $H$ and its complement. No finiteness assumption is needed.

Changing a finite generating set of a group changes individual distances, but it cannot introduce arbitrarily large relative distortion. For two finite symmetric generating sets of a group $S,T$ of the same group, setEach $S$-letter can be replaced by at most $K$ $T$-letters, and conversely each $T$-letter by at most $L$ $S$-letters. Applying these substitutions to shortest representatives of $g^{-1}h$ givesThus the identity map is a bilipschitz equivalence between the two word metrics, and hence a quasi-isometry. The constants depend on the chosen generating sets of a group.^[ref]

In the Cayley graphs of the integers example, $S=\{-1,1\}$ and $T=\{-2,-1,1,2\}$ give $K=1$ and $L=2$: a unit move costs one in either alphabet, and a move of size two costs two unit moves. This recovers $d_T\le d_S\le 2d_T$. The diamond and square in word metric balls in the integer lattice obey the same comparison, because $\max \{|m|,|n|\}\le |m|+|n|\le 2\max \{|m|,|n|\}$.

Finiteness cannot simply be dropped from this universal statement. For the additive integers, take the infinite symmetric generating set of a group $U=\mathbb{Z}\setminus \{0\}$. Every nonzero displacement is a single allowed move, so $d_U(a,b)=1$ whenever $a\ne b$, whereas $d_S(0,n)=|n|$ for $S=\{-1,1\}$. There is no finite $L$ with $d_S\le L{d}_U$: choosing an integer $n>L$ violates it. This is a counterexample to unrestricted equivalence, not a claim that every infinite generating set of a group fails; two such sets with uniformly bounded generator lengths can still satisfy the same substitution bounds.^[ref]

A group is finitely generated when it has a finite generating set of a group.

A generator of a group is a single element $g$ for which $G=⟨g⟩$: every group element is an integer power of $g$. A group admitting such an element is a cyclic group. In additive notation, every element must be an integer multiple of $g$.

Distinguish a single generator of a group from a generating set of a group. The Klein four-group needs two elements to generate it, although it has only four elements; no single element visits all four. A cyclic group need not have a unique generator of a group: $\mathbb{Z}/6\mathbb{Z}$ has $1$ and $5$, while the additive integers have $1$ and $-1$.

A group isomorphism is a bijective group homomorphism $\phi :G\to H$. It matches the entire group operation: $\phi (ab)=\phi (a)\phi (b)$, with each product taken in its own group. Its inverse is automatically a group homomorphism. A group automorphism is the special case $H=G$.^[ref]

The target is part of the claim. The doubling example under isomorphism is a group isomorphism $d:\mathbb{Z}\to 2\mathbb{Z}$, between the additive integers and the additive even integers. Exactly the same rule $d(k)=2k$, declared as $d:\mathbb{Z}\to \mathbb{Z}$, is an injective endomorphism that misses $1$ and every other odd integer, so is not a group automorphism. Changing the declared target changes surjectivity, even though it does not change the formula or its compatibility with addition.

An isomorphism lets a calculation in one model solve the same problem in another. The example under group identifies addition modulo four with function composition of rotations of a square via $k\mapsto r^k$. Through this group isomorphism, adding exponents corresponds exactly to composing motions. More generally, a group isomorphism preserves order of a group element, carries subgroups to subgroups, and carries a generating set of a group to a generating set of a group. Having equally many elements is insufficient: the permutation in group automorphism already fails compatibility on two inputs.

Why does the inverse preserve the group operation? Write $u=\phi (a)$ and $v=\phi (b)$, possible because of surjectivity. ThenThis proves the inverse condition directly. Analogous calculations work for each operation of a ring and for addition and scalar multiplication in a vector space. The argument relies on preserving those operations; it supplies no general continuity claim for the inverse of a continuous bijection between topological spaces.

The length of a word over an alphabet is the number of symbol positions in that finite string, counting repetitions. Thus $\ell (abba)=4$ and $\ell (\epsilon )=0$ for the empty word. Concatenation satisfies $\ell (uv)=\ell (u)+\ell (v)$. This differs from the word length of a group element, which minimizes over representing words using a chosen generating set of a group.

If a group has a properly discontinuous group action by isometries that is also a cocompact group action on a nonempty proper metric space that is a geodesic metric space, then the group is finitely generated and every orbit map from the word metric for any finite generating set of a group is a quasi-isometry. Finite stabilizers are permitted.

A generating set of a group from which no element can be deleted. This is inclusion-minimal, rather than necessarily having least cardinality. For finite p-groups, the Burnside basis theorem makes all such sets have the same size.

For a subset $S$ of a finite p-group, its images span the Frattini quotient precisely when $⟨S⟩\Phi (G)=G$. By the generating property proved in part (ii), this is equivalent to $⟨S⟩=G$.

If $S$ is a minimal generating set of a group, its images must be an irredundant spanning set in the vector space $G/\Phi (G)$. A linear dependence would express one image using the others; those other images would still span, so the other elements would still generate $G$, a contradiction. Thus the images form a basis. Conversely lifts of a basis generate $G$, and deleting a lift destroys spanning and hence generation.

This proves the Burnside basis theorem, including equality of sizes for every inclusion-minimal generating set of a group:Inclusion-minimal generating sets of a group need not have equal sizes in arbitrary finite groups; the elementary-abelian Frattini quotient is the essential p-group feature.

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 surviving conclusion is a quasi-isometry from a word metric for a possibly infinite generating set of a group. Finite generation cannot be asserted. Choose $x\in X$ and $R\ge 0$ so that $Gx$ is $R$-dense; the cobounded group action hypothesis provides such a choice. PutThis set is symmetric because $d_X(x,g^{-1}x)=d_X(gx,x)$. The metric geodesic subdivision argument in the next part givesIn particular $S$ generates $G$. Apply the same estimates to $g^{-1}h$ to obtain the two-sided bounds for the orbit map; its image is $R$-dense. Thus a cobounded isometric action on a metric geodesic space admits an orbit quasi-isometry for a suitable, possibly infinite, generating set. This is the cobounded orbit map lemma.

Infinite stabilizers cause no obstruction to these bounds: all their nonidentity elements belong to $S$ and have length one, although their orbit displacement is zero. The additive constant permits precisely this collapse. For example, every group acts coboundedly on a point; its complete Cayley graph for $S=G\setminus \{1\}$ is bounded, but the group need not be finitely generated.

Let $G$ have a generating set of a group with $r$ elements. Every subgroup $H$ of index $n$ gives a transitive group action of $G$ on the $n$ left cosets of $H$, hence a group homomorphism ${\rho }_H:G\to S_n$ after the cosets are labelled. Conversely, $H$ is the stabilizer subgroup of one point in this action.

A homomorphism $G\to S_n$ is determined by the images of the $r$ generators, so there are at most $(n!{)}^r$ such homomorphisms and at most $n(n!{)}^r$ point stabilizers. Therefore the finiteness of subgroups of a fixed finite index gives only finitely many subgroups of index $n$.

A group is a set $G$ equipped with a binary operation satisfying the group axioms: closure; associativity, $(gh)k=g(hk)$; an identity element $e$ with $eg=ge=g$; and an inverse element $g^{-1}$ for every $g$, with $g{g}^{-1}=g^{-1}g=e$.

A subgroup $H$ is a subset that is a group under the restricted operation. Equivalently, it is nonempty and contains $x{y}^{-1}$ whenever $x,y\in H$. A generating set of a group is a set whose elements and inverses can be multiplied in finite words to produce every group element.

The group order is its number of elements, $|G|$. The order of a group element $g$ is the least positive integer $r$ with $g^r=e$, or infinity if there is none. Lagrange's theorem states that for a finite group and subgroup $H$, $|H|$ divides $|G|$ and $|G|=[G:H]|H|$. In particular, every finite element order divides the group order.

A cyclic group is generated by one element, so every element is a power of that generator. A group isomorphism is a bijective group homomorphism: a bijection $\varphi :G\to G^{\prime }$ satisfying $\varphi (gh)=\varphi (g)\varphi (h)$ for all $g,h$.

The rank $d(G)$ of a finitely generated group is the minimum cardinality of a generating set of a group of $G$.

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.

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.