Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 3 (Topics in Infinite Groups)

Choose free group generators $a_1,\dots ,a_n$. Construct the Schreier coset graph $\Gamma$ with vertices the right cosets $Hg$ and a positively oriented edge labelled $a_k$ from $Hg$ to $Hg{a}_k$. This is well defined and connected. Its inverse edge carries $a_k^{-1}$. A word in the group generators determines a unique path from the base vertex $H$; that path closes exactly when the word represents an element of $H$.

Deleting a consecutive edge followed by its inverse corresponds exactly to free cancellation in the word. Since reduced words are unique in a free group, the resulting map from based loops modulo these cancellations to $H$ is injective as well as surjective, and respects concatenation. Thus $H$ is the fundamental group of $\Gamma$. More explicitly, choose a spanning tree $T$ and, for each unoriented edge outside $T$, choose one orientation. The tree path from the base vertex to that edge, followed by the edge and the returning tree path, defines an element of $H$. Every loop is a product of these elements and their inverses: insert the appropriate tree paths between successive edges, and tree edges contribute trivial factors. Collapsing the tree leaves one loop for each remaining edge. A product of the chosen basis loops maps to the corresponding word in these loop labels. If the original path could cancel to the constant path, its image could do so as well, since collapsing tree edges preserves every backtracking cancellation. A nonempty reduced word in the remaining labels cannot cancel to the empty word, proving that there are no further relations. These elements therefore form a free basis. This proves that $H$ is a free group without assuming the subgroup theorem.

If $i=[F_n:H]$ is finite, the graph has $i$ vertices and $ni$ unoriented edges, counted by their positive labels; loops and parallel edges count separately. A spanning tree has $i-1$ edges. The Nielsen–Schreier formula follows:For a sequence of finite indices $i_j\to \infty$, this gives the rank-to-index limit for subgroups of a free group

Finally, let $\theta :F_n\to F_n$ be an endomorphism, and put $L=\theta (F_n)$. It is generated by the $n$ images $\theta (a_k)$. If its index $d$ is finite, it is free of rank $r=1+d(n-1)$. A rank-$r$ free group needs at least $r$ group generators: its abelianization is ${\mathbb{Z}}^r$, and reduction modulo $2$ gives the $r$-dimensional vector space $(\mathbb{Z}/2\mathbb{Z}{)}^r$, which cannot be generated by fewer than $r$ vectors. ThereforeSince $n\ge 2$, this forces $d\le 1$, hence $L=F_n$. This proves that a finite-index image of a free-group endomorphism is surjective:

A group $G$ is a finitely presented group if there is an epimorphism $F(X)\to G$ from a finite-rank free group whose kernel is the normal closure of finitely many words $R$. Equivalently, $G=⟨X\mid R⟩$ with both $X$ and $R$ finite. Normal closure means that every kernel element is a finite product of conjugates of the chosen relators or their inverses.

For independence of finite presentation from finite generators, suppose $G=⟨y_1,\dots ,y_m\mid r_1,\dots ,r_s⟩$, and let $x_1,\dots ,x_k$ be any other finite generating set. Choose words $u_j(X)$ representing $y_j$ and words $v_i(Y)$ representing $x_i$. Consider the finite presentationSending each $x_i$ to its actual element gives a surjection $P\to G$. Conversely, send $y_j$ to $u_j(X)$; the substituted relators make this a homomorphism $G\to P$. Their composition on $G$ fixes every $y_j$, and their composition on $P$ fixes every $x_i$ by the extra relations. They are inverse isomorphisms. Hence finite presentability is independent of the chosen finite generating set.

For subgroups $A,B\le G$ and an isomorphism $\varphi :A\to B$, adopt the conventiona quotient of the free product $G\ast ⟨t⟩$. The generator $t$ is the stable letter of this HNN extension. Fix right coset transversals $R_{+}$ for $B\backslash G$ and $R_{-}$ for $A\backslash G$, each with $1$ representing the subgroup itself. Thus each $g$ has a unique decomposition $g=br$ with $b\in B,r\in R_{+}$, and a unique decomposition $g=ar$ with $a\in A,r\in R_{-}$.

A normal form is a stringwhere $r_i\ne 1$ whenever ${ϵ}_{i+1}=-{ϵ}_i$. The case $n=0$ is just $g_0$. The prohibition removes adjacent inverse stable letters that would cancel. The following permutation proof of HNN normal form uniqueness proves both existence and uniqueness.

Let $\mathcal{N}$ be the set of these formal strings. A base element $g$ acts by left multiplication on the initial coefficient $g_0$; call this permutation $L_g$. To define $L_t$, decompose the initial coefficient as $g_0=br$, with $b\in B,r\in R_{+}$. Unless $r=1$ and the old first stable letter is $t^{-1}$, replace the initial segment byIn the exceptional case, if the old string begins $b{t}^{-1}{r}_1\cdots$, cancel the new $t$ with that $t^{-1}$ and obtain the string beginning ${\varphi }^{-1}(b)r_1\cdots$. These operations use the defining identity $tb={\varphi }^{-1}(b)t$.

Define $L_{t^{-1}}$ similarly: write $g_0=ar$ with $a\in A,r\in R_{-}$, prepend $\varphi (a)t^{-1}r$, and cancel with an old first $t$ exactly when $r=1$. The resulting strings satisfy the normal-form conditions. These two maps are inverses. If the first operation inserts a letter, the reverse operation decomposes its new initial coefficient in the associated subgroup and cancels that letter. If the first operation cancels a letter, the reverse coset decomposition recovers the removed representative and reinserts it; an additional unwanted cancellation is excluded by the original normal-form condition.

The permutations satisfy $L_g{L}_h=L_{gh}$ andIndeed, in the decomposition $g_0=br$, multiplying $g_0$ on the left by $\varphi (a)$ leaves its representative $r$ unchanged and changes $b$ to $\varphi (a)b$. Its image under ${\varphi }^{-1}$ is $a{\varphi }^{-1}(b)$, exactly the coefficient produced by multiplying the normalized result on the left by $a$. The cancellation case has the same representative and the same conclusion. ThereforeAll defining relations of the HNN extension hold in this action, so it induces a homomorphism from $G{\ast }_{\varphi }$ into the permutations of $\mathcal{N}$.

Starting with the empty string, successive left multiplications normalize any word, and every normalization step uses a defining relation. This proves existence. If two normal words represented the same group element, their induced permutations would agree. Applied to the empty string, each normal word gives exactly its own formal string, since its representatives are already chosen and all forbidden cancellations have been excluded. Thus the strings are equal. This proves the normal form theorem for an HNN extension, including the injectivity of the base group $G\to G{\ast }_{\varphi }$.

A reduced sequence in an HNN extension is a wordwith no pinch $t^{-1}at$ for $a\in A$ and no pinch $tb{t}^{-1}$ for $b\in B$. Equivalently, if adjacent exponents are $-1,+1$, their intervening coefficient is outside $A$; if they are $+1,-1$, it is outside $B$.

Normalize such a word from right to left using the procedure above. Inductively its stable-letter length does not change. To see why, suppose the next letter has exponent $\delta$. Moving an associated-subgroup factor through it contributes to its left an element of ${\varphi }^{-1}(B)=A$ when $\delta =1$, or of $\varphi (A)=B$ when $\delta =-1$. If the preceding exponent is $-\delta$, this factor belongs to that preceding letter's pinch subgroup. Multiplying an intervening coefficient outside that subgroup by an element inside it cannot put the product inside the subgroup. Hence no cancellation occurs; when the two exponents agree, cancellation is impossible anyway. Thus a reduced word with $n\ge 1$ has a normal form still containing $n$ stable letters, which cannot be the identity normal form. This is Britton's lemma:

For the final example, write the centralizing extension asIt is generated by $a,b,t$. Suppose it were finitely presented. By the finite-generator independence already proved, the kernel of $F(a,b,t)\to E$ would be normally generated by finitely many words. Each of these kernel group generators is a finite product of conjugates of the displayed defining relators, so only finitely many indices occur in all their expressions. Choose $N$ at least as large as all of them. The full kernel would then equal the normal closure of the first $N$ relators: one inclusion follows from those expressions, and the reverse inclusion follows because all these relators already belong to the kernel.

But the group with only those first $N$ relators is the identity HNN extension over $H_N=⟨w_1,\dots ,w_N⟩$. Since the $w_j$ are a free basis of $H$, $w_{N+1}\notin H_N$. The word $t^{-1}{w}_{N+1}t{w}_{N+1}^{-1}$ is reduced in this partial extension, so Britton's lemma makes it nonidentity there. It is identity in $E$, contradicting equality of the kernels. ThereforeThis is the free-basis instance of the general fact that finite presentation of a centralizing HNN extension detects finite generation of its associated subgroup.

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.

The assumed group is a Tarski monster group. For any $g\ne 1$, its cyclic subgroup $⟨g⟩$ must be proper. Otherwise $G$ would be infinite cyclic, and $⟨g^2⟩$ would be a proper nontrivial infinite subgroup, contrary to the hypothesis. Hence $⟨g⟩$ has order $p$.

Choose $h\notin ⟨g⟩$, possible because $G$ is infinite. The subgroup $⟨g,h⟩$ contains $⟨g⟩$ strictly. If it were proper, it would have order $p$ and could not strictly contain the order-$p$ subgroup $⟨g⟩$. ThereforeIn particular, $G$ is a finitely generated group, generated by two elements.

Suppose $1\ne N⊲G$ is proper. The defining property makes $N$ cyclic of order $p$. Conjugation gives a homomorphismwhose kernel is the centralizer $C_G(N)$. The automorphism group has order $p-1$, so the kernel has finite index in $G$. It contains $N$, because $N$ is abelian. If this kernel were proper, it would be an order-$p$ subgroup, and its finite index would force $G$ itself to be finite. Therefore $C_G(N)=G$, so $N$ is central.

Choose $h\notin N$. Part (i) shows that $h$ has order $p$. Its cyclic subgroup meets $N$ trivially, since two order-$p$ subgroups with nontrivial intersection are equal. Since $N$ is central,This subgroup has order $p^2$ and is proper because $G$ is infinite, contradicting the assumed order of every proper nontrivial subgroup. Thus no such $N$ exists:Together with part (i), this proves that Tarski monster groups are two-generated and simple.

For the replacement condition, interpret “every element has order $p$” in the usual exponent-$p$ sense: every nonidentity element has order $p$, while the identity has order one. ConsiderEvery nonzero element has order $p$. It is infinite, but no finite collection generates it: the supports of finitely many group generators cover only finitely many coordinates, so a unit vector outside those coordinates is missing from their generated subgroup. Also $E$ is abelian, and one coordinate subgroup $C_p$ is a proper nontrivial normal subgroup. ThereforeThe stronger restriction on all proper subgroups, rather than just individual element orders, is what forces the conclusions in parts (i) and (ii).

The requested representation is the left coset actionIt is well defined because equal cosets remain equal after multiplication by $g$, and $\rho (g{g}^{\prime })=\rho (g)\rho (g^{\prime })$. It is transitive. For $H=1$ it is the left regular action; for general $H$ it is the associated transitive coset representation.

Its kernel is the normal core of a subgroupIndeed, $g$ fixes every $xH$ precisely when $x^{-1}gx\in H$ for every $x$. Thus $N⊲G$ and $N\le H$. If $[G:H]=i<\infty$, the image lies in the finite symmetric group $S_i$, soIn particular $N$ has finite index in $H$.

For Higman group as an amalgam of iterated HNN extensions, use the cyclic squaring presentationFirst form $L=⟨b,c\mid b^{-1}cb=c^2⟩$, an HNN extension of the infinite cyclic group $⟨c⟩$. Its base embeds, so $c$ has infinite order; $b$ has infinite order as well, since the homomorphism $L\to \mathbb{Z}$ sending $b\mapsto 1,c\mapsto 0$ is onto. Consequently $⟨b⟩$ and $⟨b^2⟩$ are isomorphic infinite cyclic subgroups. Adjoining the stable letter $a$ with $a^{-1}ba=b^2$ givesWe claim $⟨a,c⟩$ is a rank-two free group. No nonzero power of $c$ belongs to $⟨b⟩$: applying the map $b\mapsto 1,c\mapsto 0$ would make that power equal to $1$, contradicting the infinite order of $c$. Thus no nonzero power of $c$ belongs to $⟨b^2⟩$ either. Any nonempty freely reduced word in $a,c$ that contains $a$ is therefore a reduced HNN sequence: between inverse $a$ letters, a nonzero $c$ power cannot create a pinch. Britton's lemma makes it nonidentity. A word containing only $c$ is nonidentity by the base embedding. This proves the claim.

Likewiseis obtained from $⟨d,a\mid d^{-1}ad=a^2⟩$ by adjoining $c$, and its subgroup $⟨c,a⟩$ is free of rank two by the same argument. Form the amalgamated free product $P{\ast }_{⟨a,c⟩}Q$, identifying these two free subgroups generator by generator. Its presentation is exactly that of $\mathcal{H}$. The component embeddings allowed in the question show that it contains $P$, hence is infinite. It has four group generators and four relators, so it is finitely presented.

To show that this Higman group has no nontrivial finite quotient, consider any finite image and the orders of the four generator images. A relation $x^{-1}yx=y^2$ makes $y$ and $y^2$ have the same order, so every generator order is odd. Suppose one is nontrivial, and let $p$ be the smallest prime dividing any of the four orders. Choose a generator $y$ whose order is divisible by $p$, with predecessor $x$ satisfying the displayed relation. Iterating conjugation $|x|$ times givesThe multiplicative order $d$ of $2$ modulo $p$ thus divides $|x|$. Also $d\mid p-1$ and $d>1$, since $2\not\equiv 1(\mathrm{mod}p)$. Any prime divisor of $d$ is smaller than $p$ and divides the order of $x$, contradicting minimality of $p$. Hence every generator image is trivial, so every finite image is trivial. A proper finite-index subgroup would give a nontrivial transitive finite coset image, which is impossible. Therefore $\mathcal{H}$ is infinite and finitely presented, with no proper finite-index subgroups.

A maximal normal subgroup is a proper normal subgroup $M⊲G$ with no normal subgroup strictly between $M$ and $G$. To prove that finitely generated groups have maximal proper normal subgroups, fix a proper normal subgroup $K$ and order the proper normal subgroups containing $K$ by inclusion. For a chain, its union is normal and contains $K$. It is still proper: if the union were all of $G$, each element of a finite generating set would lie in some chain member, and the chain order would put all these finitely many group generators in one member. That member would equal $G$, a contradiction. The Zorn lemma therefore supplies a maximal proper normal subgroup $M$ containing $K$.

Apply this to $\mathcal{H}$ and $K=1$. By the normal-subgroup correspondence, $\mathcal{H}/M$ is nontrivial and simple. It is finitely generated as a quotient of $\mathcal{H}$. It cannot be finite, since $\mathcal{H}$ has no nontrivial finite quotients. Thus

The final assertion for arbitrary proper subgroups is false. Take $G=S_3$ and $H=⟨(12)⟩$. If a normal subgroup contained $H$, it would contain all conjugate transpositions, which generate $S_3$, so it would equal $G$. Hence this proper subgroup lies in no proper normal subgroup, and in particular in no maximal normal subgroup. The maximal-normal existence result requires the starting subgroup itself to be normal.

Let $K=\ker \theta$. The subgroup correspondence for a surjective group homomorphism sends a subgroup $B\le H$ to ${\theta }^{-1}(B)$ and a subgroup $A\le G$ containing $K$ to $\theta (A)$. Inverse images are subgroups containing $K$, and images are subgroups. Surjectivity gives $\theta ({\theta }^{-1}(B))=B$. Conversely, if $K\le A$ and $\theta (g)\in \theta (A)$, choose $a\in A$ with $\theta (g)=\theta (a)$. Then $g{a}^{-1}\in K\le A$, so $g\in A$. Thus ${\theta }^{-1}(\theta (A))=A$. These two assignments are inverse bijections.

For $C={\theta }^{-1}(B)$, the map of left coset setsis well defined and onto. If $\theta (g_1)B=\theta (g_2)B$, then $\theta (g_2^{-1}{g}_1)\in B$, so $g_2^{-1}{g}_1\in C$ and $g_{1}C=g_{2}C$. Hence it is injective, provingNo normality of $B$ is required; these are coset sets, not necessarily quotient groups.

The abelianization is $\overline{G}=G/[G,G]$, where the commutator subgroup is generated by the commutators $[x,y]=x^{-1}{y}^{-1}xy$. Conjugating a commutator gives another commutator, so this subgroup is normal. In the quotient all commutators are trivial, and therefore the quotient is abelian. If $N⊲G$ and $G/N$ is abelian, every commutator lies in $N$, so $[G,G]\le N$. Henceis well defined and satisfies $\pi =\tilde{\pi }\circ q$, where $q:G\to \overline{G}$ is the natural map. It is uniquely determined because $q$ is onto. This is the universal factorization property of abelianization.

Now write $S_i$ for the unique index-$i$ subgroup of $S$. Since conjugation preserves index, as shown in part (ii), $g{S}_i{g}^{-1}=S_i$ for every $g\in S$. Thus every $S_i$ is normal. For each prime $p$ the quotient $S/S_p$ has order $p$ and is cyclic, so there is a surjection $S\to C_p$. Being an abelian quotient, it factors through $\overline{S}$.

The abelianization is finitely generated because $S$ is. By the Fundamental theorem of finitely generated abelian groups,where $T$ is finite. If $r=0$, choose a prime $p>|T|$. The finite group $T$ could not surject onto the order-$p$ group $C_p$, contradicting the preceding surjections. Hence $r\ge 1$, and projection from a free coordinate supplies

Every subgroup of finite index in $\overline{S}$ has an inverse image of the same index in $S$, so there cannot be two different subgroups of any given finite index in $\overline{S}$. If $r\ge 2$, reduction modulo a prime $p$ in the first and second free coordinates gives two distinct surjective maps to $C_p$ with different kernels: the first basis vector lies in one kernel but not the other. Both kernels have index $p$, a contradiction. Therefore $r=1$.

If $T\ne 1$, choose a prime $p$ dividing its order. The decomposition of a finite abelian group into cyclic factors gives a surjection $\chi :T\to C_p$. In $\mathbb{Z}\oplus T$, the mapshave distinct index-$p$ kernels, again distinguished by $(1,0)$. This is another contradiction, so $T=1$. We have proved uniqueness of finite-index subgroups forces infinite cyclic abelianization:

A residually finite group is one in which every nonidentity element has nonidentity image under some homomorphism to a finite group. Equivalently, the intersection of all finite-index normal subgroups is trivial: a finite separating map has such a kernel, and conversely quotienting by such a normal subgroup gives a finite separating map.

Choose the surjection $\sigma :S\to \overline{S}\cong \mathbb{Z}$. For each positive integer $i$, the subgroup ${\sigma }^{-1}(i\mathbb{Z})$ has index $i$ and hence equals $S_i$. Its kernel $[S,S]$ is therefore contained in every finite-index subgroup. Indeed,If $S$ is residually finite, this intersection is trivial. Thus $[S,S]=1$ and $\sigma$ is an isomorphism. The residually finite group with one subgroup of each finite index is infinite cyclic conclusion isResidual finiteness is used exactly to eliminate the commutator subgroup that all finite quotients would otherwise fail to detect.

For an arbitrary subgroup $A$, the full inverse image of its image is $AK$, not generally $A$. In fact, $\theta (g)\in \theta (A)$ exactly when $g=ak$ for some $a\in A,k\in K$. Since $K$ is normal, $AK$ is a subgroup. By the subgroup correspondence for a surjective group homomorphism,Multiplicativity of finite index givesThe map $k(K\cap A)\mapsto kA$ bijects the left cosets of $K\cap A$ in $K$ with those of $A$ in $AK$, since $AK=KA$. ConsequentlyThis need not equal $i$. For example, take $\theta :\mathbb{Z}\to \mathbb{Z}/2\mathbb{Z}$ to be reduction modulo $2$ and $A=3\mathbb{Z}$. Then $[G:A]=3$, but $\theta (A)=H$ has index one. If $A$ contains $K$, however, the denominator is one andFor finite $i$, equality holds exactly when $K\le A$.

Conjugation ${\alpha }_g(x)=gx{g}^{-1}$ is an automorphism of $G$ sending $A$ to $gA{g}^{-1}$. It induces the bijectionbetween their left coset sets; conjugation by $g^{-1}$ gives the inverse. ThereforeThis holds whether or not $A$ is normal, and whether the index is finite or infinite.