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If $H$ is a finite-index subgroup of a finitely generated group $G$, then $H$ is finitely generated and its inclusion into $G$ is a quasi-isometry. Finite generation follows from Schreier's lemma, while finitely many coset representatives give coarse surjectivity and uniformly bounded distortion.

A finitely generated abelian group whose elements all have finite order is finite. Induct on derived length for a finitely generated soluble group with this property. Its abelianization is finite, so its commutator subgroup has finite index and is finitely generated by Schreier's lemma. That subgroup is soluble of smaller derived length and is still a torsion group, hence finite by induction. The ambient group is a finite extension of it and is finite.

For $1\to N\to G\to Q\to 1$ with $N,Q$ locally finite, take a finitely generated subgroup $L\le G$. Its image in $Q$ is finite, so $L\cap N$ has finite index in $L$ and is finitely generated by Schreier's lemma. Local finiteness of $N$ makes that intersection finite. Thus $L$ itself is finite, proving local finiteness of $G$.

// Target: algebra.bigb

Choose representatives $T$ of the right cosets $Hg$, including $1$, and let $M={\max }_{t\in T}|t{|}_A$. This is finite because the index is finite. For $t\in T$ and $a\in A^{\pm 1}$, let $\overline{ta}\in T$ represent $Hta$. The elementsform a finite generating set $S$ for $H$, after adjoining inverses and discarding identities. To see generation rather than merely quote Schreier's lemma, write $h=a_1\cdots a_n$ and let $t_i$ represent $H{a}_1\cdots a_i$. Then $t_0=t_n=1$ andEach factor is a Schreier generator. For a shortest $A$-word this gives $|h{|}_S\le |h{|}_A$. Conversely every Schreier generator has $A$-length at most $2M+1$, so $|h{|}_A\le (2M+1)|h{|}_S$. Hence the inclusion $H\hookrightarrow G$ obeysEvery $g\in G$ can be written $g=ht$, and $d_A(g,h)=|t{|}_A\le M$. Thus the inclusion is coarsely onto and is a finite-index subgroup quasi-isometry.

Only finite generation of $G$ was needed for the metric argument. Under the given finite-presentation hypothesis, $H$ is also finitely presented: the corresponding finite-sheeted cover of a finite presentation complex has finitely many cells, and a spanning-tree reduction of its one-skeleton gives a finite group presentation. Therefore it belongs to the class of groups used in part (ii).

For a finitely generated group $G=⟨x_1,\dots ,x_r⟩$, every quotient group is generated by the images of the $x_i$. Thus finite generation passes to quotients.

For a group extension $1\to N\to G\to Q\to 1$, suppose $N$ and $Q$ are finitely generated. Choose finite group generators of $N$ and lifts to $G$ of finite group generators of $Q$. For any $g\in G$, a word in the lifts has the same image as $g$ in $Q$, so their difference belongs to $N$. Thus the two finite sets together generate $G$, proving preservation under extensions.

For a subgroup $H\le G$ of finite index, choose representatives $T$ of the right cosets $H\backslash G$, with $1$ representing $H$, and a finite symmetric generating set $X$ for $G$. Write $\overline{g}$ for the chosen representative of $Hg$. The finitely many Schreier generatorsbelong to $H$. If $h=x_1\cdots x_k\in H$, set $t_j=\overline{x_1\cdots x_j}$, so $t_0=t_k=1$. ThenEach factor is a listed generator because $H{x}_1\cdots x_j=H{t}_{j-1}{x}_j$. Therefore $H$ is finitely generated. This proves Schreier's lemma directly.

An infinite locally finite group isIts elements are binary sequences of finite support, with coordinatewise addition. It is infinite because the unit vectors are distinct. Any finite collection of elements has its supports contained in one finite set of coordinates, and the subgroup it generates lies in the corresponding finite product of copies of $C_2$. Hence every finitely generated subgroup is finite.

Local finiteness passes to subgroups because a finitely generated subgroup of a subgroup is also one of the ambient group. It passes to quotients as well: lift finitely many group generators of a subgroup in the quotient; the subgroup generated by their lifts is finite, and its image is exactly the desired subgroup.

For local finiteness is closed under extensions, suppose $N$ and $Q=G/N$ are locally finite, and take a finitely generated subgroup $K\le G$. Its image $\overline{K}\le Q$ is finitely generated, hence finite. The map $K\to \overline{K}$ has kernel $K\cap N$, soThe finite-index result already proved makes $K\cap N$ finitely generated. Local finiteness of $N$ makes it finite, and thenSince $K$ was arbitrary, $G$ is locally finite. Thus finite generation is preserved by quotients, extensions and finite-index subgroups; local finiteness is preserved by subgroups, quotients and extensions.

We construct an infinite residually finite torsion group using the Golod-Shafarevich infinitude criterion for pro-p groups. Its useful form says that a pro-$p$ presentation on $d$ generators with relators $r_i$ gives an infinite pro-p group wheneverHere $D$ is the Magnus degree of a free pro-p group element: map each generator to $1+X_j$ in a noncommutative power-series algebra over ${\mathbb{F}}_p$, and take the lowest degree of $w-1$. The criterion permits infinitely many relators with a convergent weight sum. It is a consequence of the Golod-Shafarevich inequality for the completed group algebra; see Ershov's survey, Theorem 3.1.

Take $p=2$, $d=2$, and $\tau =3/4$. Let $F$ be the free pro-p group on $a,b$, and enumerate all ordinary finite words $w_1,w_2,\dots$ in these generators and their inverses. Choose integers $e_i$ sufficiently large thatForm the quotient pro-p groupwhere the bar means the closed normal closure. If the Magnus image of $w_i$ is $1+v_i$, characteristic two givesfor nonidentity $w_i$; identity relators contribute zero weight. It follows thatThus the Golod-Shafarevich infinitude criterion for pro-p groups proves that $P$ is infinite.

Let $G=⟨\bar{a},\bar{b}⟩$ be the abstract subgroup generated by the two images inside $P$. This subgroup is dense because the two images topologically generate the quotient, and it is infinite: a finite subgroup of a Hausdorff profinite group is closed, so could not be dense in infinite $P$. Every element of $G$ is the image of some ordinary word $w_i$, and the imposed relation makes its order divide $2^{e_i}$. Therefore $G$ is a two-generated torsion group. Finally $P$ is an inverse limit of finite 2-groups. Any nonidentity element of $G$ is nonidentity in $P$ and hence has a nonidentity image in one of those finite groups. Restricting that map to $G$ proves that it is a residually finite group. We have obtainedThe abstract subgroup $G$ is essential here: the argument establishes finite order for ordinary words, not for every topological limit in the ambient pro-p group.

There is no surjective homomorphism from this group to an infinite soluble group. In fact finitely generated soluble torsion groups are finite. To prove this, let $S$ be finitely generated, soluble and torsion, and induct on its derived length. In the abelian case, finitely many generators of finite orders generate a finite group. In general the abelianization $S/S^{\prime }$ is a finitely generated abelian torsion group, hence finite. Its kernel $S^{\prime }$ therefore has finite index and is finitely generated by Schreier's lemma. Explicitly, for a finite coset transversal $T$ containing $1$ and a finite symmetric generating set $X$, the finitely many elements $tx{\overline{tx}}^{-1}$ generate the finite-index subgroup, where the bar denotes the chosen representative of the coset. Now $S^{\prime }$ is torsion and has smaller derived length, so induction makes it finite. Hence $S$ is finite as well. Every quotient of $G$ is finitely generated and torsion, so any soluble quotient must be finite.

Let $r=d(G)$ be the rank of a group, let $S$ be a generating set of size $r$, and suppose $[G:H]=d$. Choose a Schreier transversal adapted to a spanning tree in the Schreier coset graph. By Schreier's lemma, $H$ is generated by the elementsThere are $dr$ candidates. The $d-1$ oriented edges in the spanning tree give trivial candidates, leaving at most $dr-(d-1)$. This proves the Schreier index-rank inequality

Identify $G$ with the finite-index subgroup $f(G)\le H$. If $S$ is a finite generating set for $H$ and $T$ is a finite set of right-coset representatives, Schreier's lemma gives a finite generating set for $f(G)$, and hence for $G$.

Equip both groups with word metrics from finite generating sets. The inclusion is Lipschitz because each generator of $G$ has bounded length in $H$. Conversely, rewriting a word in $S$ by tracking its cosets through the finite set $T$ expresses an element of $G$ as a word of length bounded linearly in its $H$-length. Finally, every element of $H$ lies within the maximum word length of an element of $T$ from $f(G)$. Thus the inclusion is a finite-index subgroup quasi-isometry, and composing it with the isomorphism $f:G\to f(G)$ proves that $G$ and $H$ are quasi-isometric.

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.