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A group $G$ is residually finite if, for every $1\ne g\in G$, there are a finite group $Q$ and a homomorphism $\varphi :G\to Q$ such that $\varphi (g)\ne 1$.

Let $F(S)$ be a free group and let $1\ne w\in F(S)$ be a reduced word. In the rose with one oriented loop for each $s\in S$, the word $w$ determines a nonclosed reduced path from a chosen vertex in the universal covering tree. Its finite image path can be completed to a finite covering graph of the rose: for each label, pair the still unmatched incoming and outgoing edge germs, adding finitely many vertices if necessary. The lift of $w$ remains nonclosed in this finite cover.

Let $H\le F(S)$ be the finite-index subgroup represented by this based cover. Then $w\notin H$. The action of $F(S)$ on the finite set of right cosets $H\backslash F(S)$ gives a homomorphism to a finite symmetric group, and $w$ does not fix the coset $H$. Thus $w$ survives in a finite quotient. Since $w$ was arbitrary, every free group is residually finite.

Put $Q=G/H$ and write $M^H$ for the invariant submodule. The five-term exact sequence in group cohomology associated with the Lyndon–Hochschild–Serre spectral sequence isThe inflation map in group cohomology composes a cocycle on $Q$ with the quotient homomorphism $G\to Q$. The restriction map in group cohomology restricts a cocycle from $G$ to $H$. The quotient action on the middle term is, for $q=gH$ and a one-cocycle $f$,This is independent of the lift and of the representative at the level of cohomology. Finally, the transgression in group cohomology extends a $Q$-invariant class on $H$ to a one-cochain on $G$; its coboundary is $H$-basic and descends to the two-cocycle on $Q$ representing $d_2[f]$. Changing the extension changes that cocycle by a group coboundary.

For the application, choose free generators $x_1,\dots ,x_n$ of $F$ and normal generators $r_1,\dots ,r_n$ of $R$. Since a finite nonabelian simple group $K$ is a perfect group, its abelianization is zero. The five-term sequence for $1\to R\to F\to K\to 1$ with trivial coefficients containsThe first term is zero because $K$ is finite, and the last term is zero because a free group has cohomological dimension one. It remains to prove that restriction is surjective.

The invariant submodule of homomorphisms $R\to \mathbb{Z}$ is exactlyThe images of the relators $r_1,\dots ,r_n$ generate $R/[F,R]$, so such a homomorphism $\lambda$ is determined by the integer vector $b=(\lambda (r_1),\dots ,\lambda (r_n){)}^T$. Let $A$ be the relator exponent-sum matrix. This square integer matrix presents $K^{\mathrm{ab}}$, which is zero, so $A$ is a unimodular matrix. There is therefore an integer vector $v$ satisfying $Av=b$. Define $\psi :F\to \mathbb{Z}$ by assigning to $x_i$ the $i$th entry of $v$. The definition of $A$ gives $\psi (r_j)=\lambda (r_j)$ for every $j$. Since the relator images generate $R/[F,R]$, the restriction of $\psi$ to $R$ equals $\lambda$. Restriction is surjective, exactness now gives

For a group presentation $\mathcal{P}=⟨X\mid R⟩$ with finitely many generators and a prime number $p$, defineThis uses the unshifted convention. Identity relators have zero weight, and a divergent sum gives value $-\infty$. Some literature subtracts one instead; inequalities must be shifted accordingly. Taking roots in the free group matters: roots appearing only after passing to the quotient do not change the relator weight. The p-rank of a group bounds this quantity above, and the index-p rewriting bound for p-deficiency gives the infinitude criterion p-deficiency at least one implies infinitude.

For finite-index subgroups of a rank-$n$ free group, the Nielsen–Schreier formula gives the displayed quotient. Along any sequence whose finite indices tend to infinity, the quotient tends to $n-1$. Subtracting one from the rank before dividing gives the constant rank gradient $n-1$.

// Target: geometry-and-topology.bigb

A reduced word contains no adjacent generator and inverse-generator pair. Cancel such pairs to obtain the unique reduced representative in a free group. One proof constructs the free group itself from reduced words, with multiplication given by concatenation followed by cancellation; its action by these operations on words distinguishes distinct reduced representatives.

A relator is a word in the generators of a free group that is imposed as the identity in a group presentation.

Every finitely generated free group is a residually finite group. For $F_2$, the ping-pong lemma embeds it into ${\mathrm{SL}}_2(\mathbb{Z})$, and reduction modulo a prime in an integral matrix group separates each nonidentity element in a finite quotient.

If $H$ has finite index $d$ in a finitely generated group $G$, then Schreier's lemma givesEquality holds when $G$ is a free group of finite rank.

For $S\subseteq {\mathbb{N}}_{>0}$, form the Right-angled Artin group on generators $b_i$ indexed by $i\in \mathbb{Z}$, imposing $[b_i,b_j]=1$ precisely when $|i-j|\in S$. Translation $b_i\mapsto b_{i+1}$ is an automorphism. Its semidirect product with an infinite cyclic group generated by $t$ is countable and generated by $b_0,t$. If $n\notin S$, killing all $b_i$ except $b_0,b_n$ retracts the graph group onto the rank-two free group, so their commutator is nonidentity. This proves the displayed equivalence. For non-recursively-enumerable $S$, enumerable identities of finitely generated subgroups rules out an embedding of this two-generated group in any finitely presented group.

A group has integral cohomological dimension at most one precisely when it is a free group, including the trivial free group in dimension zero. This result concerns integral coefficients; a rational dimension-one statement has different conclusions.

Consider a connected assembly of $n$ closed discs formed by identifying $e$ pairs of boundary arcs, each arc used once and all arcs in a given tile having disjoint endpoints. Its gluing graph has one vertex per tile and one edge per paired arc. A spanning tree joins the initial discs into a single closed disc; each of the remaining $e-n+1$ pairings attaches a band. This is a thickening of the gluing graph, so it is homotopy equivalent to that graph. Thus its Euler characteristic is $n-e$ and its fundamental group is the free group $F_{e-n+1}$. If the gluing graph is a tree, the assembled surface is itself a closed disc.

Suppose each pairing preserves the positive boundary parameter of the reference discs. A global orientation must then assign opposite orientation signs to adjacent tiles, because their induced boundary orientations on an interior seam must oppose one another. Such signs exist exactly when the gluing graph is bipartite. An odd gluing cycle therefore proves that the assembly is a nonorientable surface. A Dirichlet boundary condition or a Neumann boundary condition on an unpaired arc has no effect on this topological calculation.

A group is a torsion group when every element is a torsion element, meaning that each element has finite order. The orders need not have a common finite bound. A torsion group has no nontrivial torsion-free group as a subgroup; in particular it cannot contain a nonabelian free group. Infinite finitely generated groups of this kind can be constructed using a torsion group construction by p-power relators.

Enumerate all nonidentity words $w_i$ in the rank-two free group and impose relations $w_i^{p^{i+2}}=1$. The resulting two-generated group hasIt is infinite by p-deficiency at least one implies infinitude. Each element has order a power of $p$, so it is a torsion group. The infinitely many relators are essential to this particular construction; finite generation does not imply a finite group presentation.

These upper and lower quantities use $d(H)$, without subtracting one. For a free group $F_n$ with $n\ge 1$, the Nielsen–Schreier formula gives $d(H)/[F_n:H]=n-1+1/[F_n:H]$. The upper value is $n$, attained at the whole group, and the lower value is $n-1$, approached by kernels of maps onto cyclic groups of arbitrarily large finite order. The frequently used shifted rank gradient replaces $d(H)$ by $d(H)-1$ and therefore gives different upper values.

A group is virtually free when it has a free group as a finite-index subgroup. Every finitely generated virtually free group is quasi-isometric to a tree.

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.