Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 10 (Topics in Infinite Groups)

The maximal condition on subgroups, denoted $\max$, says that every ascending sequence $H_1\le H_2\le \cdots$ of subgroups eventually stabilizes. Suppose a subgroup $H$ were not finitely generated. Starting with the trivial subgroup, repeatedly choose an element of $H$ outside the subgroup generated so far. This produces a strictly ascending sequence, contradicting $\max$.

Conversely, suppose every subgroup is a finitely generated group. The union $H={\bigcup }_i{H}_i$ of an ascending sequence is a subgroup. A finite generating set of $H$ lies in a single $H_m$, since the sequence is ascending. Thus $H=H_m$, and $H_i=H_m$ for all $i\ge m$. We have proved

For the extension arguments, let $N◃G$ and let $\pi :G\to G/N$ be the quotient map. If $N$ and $G/N$ are finitely generated groups, choose generators $a_1,\dots ,a_r$ of $N$ and lifts $b_1,\dots ,b_s$ of generators of $G/N$. For every $g\in G$, its image is a word $w(\pi (b_1),\dots ,\pi (b_s))$, so $w(b_1,\dots ,b_s{)}^{-1}g\in N$. ThereforeHere $d$ denotes the minimal number of generators. If instead $N$ and $G/N$ satisfy the maximal condition on subgroups, take any $H\le G$. Its intersection $H\cap N$ and its image $\pi (H)$ are finitely generated by the condition in those two groups. The exact sequence $1\to H\cap N\to H\to \pi (H)\to 1$ and the preceding argument make $H$ finitely generated. The equivalence already proved now gives $\max$ for $G$. Hence both finite generation and the maximal condition are preserved by group extensions.

An infinitely generated group of finite Prüfer rank is the additive group $\mathbb{Q}$. A finite set of rational numbers lies in $D^{-1}\mathbb{Z}$ for a common denominator $D$, and its generated subgroup is cyclic: after multiplying by $D$, use the greatest common divisor of the integer numerators. Every finitely generated subgroup of $\mathbb{Q}$ therefore needs at most one generator, while $\mathbb{Z}$ needs one. But no finitely generated subgroup contains $1/(2D)$ once its elements all lie in $D^{-1}\mathbb{Z}$. Thus

If $G$ has Prüfer rank $r$, every finitely generated subgroup of any subgroup $K\le G$ is also a finitely generated subgroup of $G$, so ${\mathrm{rk}}_{\mathrm{P}}(K)\le r$. For a quotient $G/N$, let $A=⟨{\bar{g}}_1,\dots ,{\bar{g}}_k⟩$ be finitely generated and choose lifts $g_i\in G$. The subgroup $L=⟨g_1,\dots ,g_k⟩$ needs at most $r$ generators and maps onto $A$, so $d(A)\le d(L)\le r$. This proves the rank bound for every subgroup and quotient.

Now suppose $N$ and $G/N$ have Prüfer ranks $r$ and $s$. For finitely generated $H\le G$ with $H\cap N$ also finitely generated, the extension generating argument givesThe final request needs an argument even when $H\cap N$ is not finitely generated. Write $H=⟨h_1,\dots ,h_k⟩$. Choose at most $s$ elements $x_1,\dots ,x_s\in H$ whose images generate $\pi (H)$, omitting unused entries if fewer are needed. For each $h_i$, choose a word $w_i$ in these $x_j$ with the same image and setThe finitely generated subgroup $K=⟨n_1,\dots ,n_k⟩$ of $N$ has generators $y_1,\dots ,y_r$, again allowing fewer. Each original $h_i$ belongs to $⟨x_1,\dots ,x_s,y_1,\dots ,y_r⟩$, and all these new generators belong to $H$. Thus $d(H)\le r+s$ without requiring finite generation of the full intersection. Taking the supremum proves the Prüfer rank bound for extensions:

The free product $G_1\ast G_2$ is a group with homomorphisms from $G_1,G_2$ such that any pair of homomorphisms $G_i\to K$ extends uniquely to a homomorphism $G_1\ast G_2\to K$. Concretely its elements are reduced words with nonidentity syllables alternating between the factors. Multiply by concatenating and then combining adjacent syllables from the same factor, deleting identities. The empty word is the identity, and this normal form theorem for a free product also makes the factor maps injective.

The Klein combination theorem, in its ping-pong form, is as follows. Suppose two nontrivial subgroups $A,B$ act on a set $X$, with disjoint nonempty subsets $X_A,X_B$, such thatIf at least one factor has at least three elements, then $⟨A,B⟩\cong A\ast B$ through the natural factor maps. The cardinality condition is needed in this version: two order-two groups can have the same image swapping the two domains.

Here is the proof. Assume $|A|\ge 3$. A reduced word beginning and ending with an $A$-syllable sends $X_B$ into $X_A$ by successive application of the displayed inclusions, so cannot act as the identity. A word beginning and ending in $B$ becomes one of the preceding type on conjugating by any nonidentity $a\in A$. If it begins in $A$ and ends in $B$, choose $a\in A$ different from both $1$ and its first syllable. The word $a^{-1}wa$ remains reduced after combining its first two syllables and begins and ends in $A$, so again it is nonidentity. If it begins in $B$ and ends in $A$, apply this last argument to its inverse. Every nonempty reduced word is covered, so the natural homomorphism $A\ast B\to ⟨A,B⟩$ has trivial kernel. This proves the Klein combination theorem.

Apply it to the two cyclic groups generated by the given Möbius transformations. On $X=\mathbb{R}\cup \{\infty \}$ takeTheir powers areFor $m\ne 0$ and $|x|<1$, $|x+2m|\ge 2|m|-|x|>1$, so $f^m{X}_B\subseteq X_A$. For $x\in X_A$, put $1/\infty =0$; thenso $g^m{X}_A\subseteq X_B$. The powers displayed also show that both transformations have infinite order. The Klein combination theorem consequently gives

For the finite-index formula, realize the free group $F_n$ as the fundamental group of the rose with $n$ edges. A subgroup $H$ of index $i$ corresponds to a connected $i$-sheeted covering graph. Explicitly, its vertices are right cosets $Hx$ and for each basis generator $s_j$ there is a directed edge $Hx\to Hx{s}_j$ projecting to the corresponding rose edge. This graph is connected because basis words reach every coset, and the based loops correspond precisely to words in $H$.

The covering graph has $i$ vertices and $ni$ geometric edges: each positively oriented rose edge has one lift from each vertex. A maximal tree has $i-1$ edges, and the complementary edges freely generate its fundamental group. Thus $H$ is a free group with the Nielsen–Schreier formulaThis uses the connected cover associated with $H$; no claim that arbitrary disconnected covers are connected is needed.

Finally the specified basis of $F_n$ gives $d(F_n)\le n$. Its abelianization is ${\mathbb{Z}}^n$, by the universal property of the free group, and reduction modulo two gives a quotient $(\mathbb{Z}/2\mathbb{Z}{)}^n$. Images of any generating set must span this $n$-dimensional vector space, so it contains at least $n$ members. Therefore $\boxed{d(F_n)=n}$, and the same argument applies to the free group $H$ above.

Using exactly the unshifted definition printed in the paper, the unshifted rank gradients are determined byThe supremum is $n$, attained for $H=F_n$. For $n\ge 1$, mapping one basis generator onto a generator of $\mathbb{Z}/i\mathbb{Z}$ and the others to zero provides subgroups of every positive index $i$. Hence the infimum is $n-1$, approached as $i\to \infty$:For the trivial free group $F_0$, both values are zero. Subtracting one from $d(H)$ would change the upper answer; that is a different convention from the one requested here.

A finitely presented group is one isomorphic to $F(X)/⟨⟨R⟩⟩$, where $X$ and $R$ are finite, $F(X)$ is the free group on $X$, and the double brackets mean normal closure. Equivalently, there is a surjection from a free group of finite rank onto $G$ whose kernel is normally generated by finitely many words. The universal property of the free group extends choices of generator images uniquely, and quotienting imposes the chosen relations.

For the semidirect product, choose finite group presentationsFor each $i,j$, choose an $a$-word $w_{ji}$ representing $\varphi (b_j)(a_i)$. We claim that a finite presentation of a semidirect product isAll relations hold in the desired semidirect product, so the presented group maps onto it. We must also show there are no extra elements in the kernel. For inverse conjugations, choose an $a$-word $v_{ji}$ representing $\varphi (b_j{)}^{-1}(a_i)$. Conjugating this word by $b_j$ using the displayed relations gives an $a$-word representing $\varphi (b_j)\varphi (b_j{)}^{-1}(a_i)=a_i$, so $R$ proves $b_j^{-1}{a}_i{b}_j=v_{ji}$. Thus we can move both positive and negative $b$-letters to the right, rewriting every word as $u(a)v(b)$. If its image in $N⋊H$ is trivial, projection to $H$ makes $v$ trivial using $S$; its remaining image in $N$ then makes $u$ trivial using $R$. The kernel is trivial, proving the claim and finite presentability.

For a poly-infinite cyclic group, write its subnormal series as $G=G_0▹G_1▹\cdots ▹G_n=1$ with infinite cyclic factors. Given a subgroup $K\le G$, intersect the series with $K$. The homomorphism $K\cap G_i\to G_i/G_{i+1}$ has kernel $K\cap G_{i+1}$, soEach factor is trivial or infinite cyclic; deleting repetitions gives a poly-infinite cyclic series for $K$. For an extension $1\to N\to E\to Q\to 1$ with $N,Q$ poly-infinite cyclic groups, lift the series of $Q$ to its full preimages in $E$ and append the series of $N$. The lifted quotient factors are those of $Q$, and the lower factors are those of $N$. Normality at every consecutive step is preserved. This proves closure under both subgroups and extensions.

Finite presentability follows by induction on the length of a poly-infinite cyclic series. The trivial group is finitely presented. For a nontrivial group, its first proper term $G_1$ has a shorter series and $G/G_1\cong \mathbb{Z}$. Lift $1\in \mathbb{Z}$ to an element $t\in G$. Its powers give an injective section of the quotient map, since $t^k$ maps to $k$. Hence $G\cong G_1⋊⟨t⟩$. By induction $G_1$ is finitely presented, and $⟨t⟩\cong \mathbb{Z}$ has a finite presentation. The finite presentation of a semidirect product proved above completes the induction. Thus every poly-infinite cyclic group is finitely presented.

For the final example, take the dyadic affine Baumslag-Solitar groupwhere the generator of $\mathbb{Z}$ acts on the additive group $\mathbb{Z}[1/2]$ by multiplication by two. Here is a verification of this identification, rather than an assumption about the presentation. Put $a_i=t^{-i}a{t}^i$ for $i\ge 0$. The relation gives $a_{i+1}^2=a_i$, soTheir union $A$ is abelian and is normal: conjugation by $t^{-1}$ shifts $a_i$ to $a_{i+1}$, while conjugation by $t$ shifts $a_i$ to $a_{i-1}$ for $i\ge 1$ and sends $a_0$ to $a_0^2$. Every word can consequently be written $a_i^m{t}^k$ for some $i\ge 0$ and $m,k\in \mathbb{Z}$.

Let $a$ act on the rational line by $x\mapsto x+1$ and $t$ by $x\mapsto 2x$. These affine transformations satisfy the relation, and $a_i^m{t}^k$ acts by $x\mapsto 2^{k}x+m/2^i$. It is the identity only when $k=m=0$. Thus the action is faithful and identifies $A$ with all dyadic translations. We have the required seriesBoth factors are torsion-free abelian groups, and $G$ is visibly finitely presented. But $\mathbb{Z}[1/2]$ is not finitely generated, since finite sets have bounded powers of two in their denominators. Every poly-infinite cyclic group has the maximal condition on subgroups: its factors $\mathbb{Z}$ have that condition, and Question 1 proves closure under extensions. Therefore $G$ cannot be poly-infinite cyclic. This gives a finitely presented counterexample to replacing infinite cyclic factors by arbitrary torsion-free abelian factors.

Suppose $G$ is generated by $d$ elements. An index of a subgroup $n$ gives a transitive action of $G$ on $n$ cosets. After labelling them by $1,\dots ,n$, this is a homomorphism $G\to S_n$, and the subgroup is the stabilizer of one label. There are at most $(n!{)}^d$ homomorphisms, since images of the generators determine the map, and at most $n$ point stabilizers for each map. Thus there are finitely many subgroups of each specified finite index, with the coarse bound $n(n!{)}^d$.

The finite residual isIt is equivalently the intersection of the kernels of all homomorphisms to finite groups. Intersecting all finite-index subgroups gives the same subgroup, since each contains its normal core, a normal subgroup of finite index. A residually finite group is one with $R_G=1$, equivalently every nonidentity element survives in a finite image.

A Hopfian group is one whose surjective endomorphisms are all injective, hence automorphisms. Let $\theta :G\to G$ be surjective, retaining the finite-generation hypothesis. For each $n$, pulling back by $\theta$ sends a normal subgroup of index $n$ to another normal subgroup of the same index. It is injective on this finite set: equality of two preimages implies equality of the subgroups because $\theta$ is surjective. Therefore it is a permutation. For every finite-index normal subgroup $N$, some $m\ge 1$ satisfiesIt follows that $\ker \theta \subseteq \ker {\theta }^m\subseteq N$. Intersecting over $N$ proves finite residual contains surjective endomorphism kernels:If $G$ is also a residually finite group, this forces every surjective endomorphism to have trivial kernel, proving that a finitely generated residually finite group is Hopfian.

For the specified Baumslag-Solitar group, use the convention $[x,y]=xy{x}^{-1}{y}^{-1}$ and writeIn the abelianization, the defining relation gives $2a=3a$, so the image of $a$ is zero, while $t$ generates an unrestricted infinite cyclic group. Thus $G/G^{\prime }\cong \mathbb{Z}$ through the exponent sum of $t$. The kernel is generated by all $a_i$: any word can be rewritten as a product of conjugates of $a$ followed by the final power of $t$. Alternatively $a=[t,a^2]\in G^{\prime }$, so all $a_i\in G^{\prime }$, while killing them leaves a cyclic quotient. Therefore

Define $\theta (a)=a^2$ and $\theta (t)=t$. This respects the defining relation becauseThe image contains $a^2$, $t$, and $t{a}^2{t}^{-1}=a^3$, hence contains $a=a^3(a^2{)}^{-1}$ and is all of $G$. It is not injective. Indeedmaps to $[a^3,a^2]=1$. But it is a reduced word in the HNN extension with associated subgroups $⟨a^2⟩$ and $⟨a^3⟩$. Its potential subwords $t{a}^{\pm 1}{t}^{-1}$ do not have even exponent, and its potential subword $t^{-1}at$ does not have exponent divisible by three. It has no pinch in an HNN extension, so Britton's lemma gives $c\ne 1$. We have explicitly exhibited a surjective, noninjective endomorphism.

It remains to treat the whole second commutator subgroup, not just this one kernel element. SetThese kernels form an increasing sequence of normal subgroups, so $K$ is a normal subgroup. Every iterate is surjective, and the preceding finite residual argument gives $K\subseteq R_G$.

Conjugating the defining relation gives $a_{i+1}^2=a_i^3$ for every integer $i$. Repeated substitution gives, for $j>i$,Also ${\theta }^k(a_i)=a_i^{2^k}$. Thus when $k\ge j-i$,Every pair of the generating elements of $G^{\prime }$ therefore commutes modulo $K$. The image of $G^{\prime }$ in $G/K$ is abelian, which means $G^{\prime \prime }\subseteq K$. This proves the full statement using iterated endomorphism kernels in BS(2,3):

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.