Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 104 (Infinite Groups)

Because the generating sets are disjoint, the free product has the presentationThis is immediate from the universal property of a group presentation: a map out of this presented group is exactly a pair of homomorphisms from $G$ and $H$ to the target group.

A reduced word is either the empty word or a productin which every syllable $u_i$ is a nonidentity element of $G$ or $H$, and consecutive syllables belong to different factors.

Let $\mathcal{W}$ be the set of reduced words. Each $g\in G$ acts on the right of $\mathcal{W}$: if the last syllable lies in $H$, append $g$; if it lies in $G$, multiply it by $g$ and delete it when the product is the identity. Define the action of each $h\in H$ analogously. These rules give genuine actions of the two factors by permutations of $\mathcal{W}$, hence an action of the free group $F(X\bigsqcup Y)$. Every relation in $R\bigsqcup S$ acts trivially, so the action factors through the displayed presentation of $G\ast H$.

The element represented by a reduced word $w$ sends the empty word to $w$. Therefore two reduced words representing the same element induce the same permutation and have the same value on the empty word. They must be identical. This proves the normal form theorem for a free product.

Let $w$ be a nonempty reduced word. If its first syllable belongs to $G$, choose any nonidentity $h\in H$. The reduced form of $hw$ begins with an $H$-syllable, whereas that of $wh$ begins with the original $G$-syllable; reduction at the right end cannot change the first syllable. Hence $hw\ne wh$. The same argument with a nonidentity $g\in G$ applies when $w$ begins in $H$. No nonidentity element is therefore central, and

Choose distinct nonidentity elements $g_1,g_2\in G$ and $h_1,h_2\in H$, and putExpand a freely reduced word in $u^{\pm 1},v^{\pm 1}$. At a boundary where two syllables from one factor meet, their product is one of $g_1^{-1}{g}_2$, $g_2^{-1}{g}_1$, $h_1{h}_2^{-1}$, or $h_2{h}_1^{-1}$, all nonidentity by the choices above. Every other boundary already alternates between the factors. Thus the expansion reduces to a nonempty reduced word in $G\ast H$ and cannot represent the identity. The homomorphism from the rank-two free group sending its free generators to $u,v$ is injective, so

The Integer Heisenberg groupis generated by the matrices with $(a,b,c)=(1,0,0)$ and $(0,1,0)$. Their group commutator is the nonidentity central matrix with $(a,b,c)=(0,0,1)$. Thus it is nonabelian and nilpotent of class two, while being finitely generated.

The countable direct sumis abelian and therefore nilpotent of class one. It is not finitely generated, whereas every polycyclic group is finitely generated. Hence it is nilpotent but not polycyclic.

The lamplighter groupis generated by one lamp switch and one translation. It is metabelian, hence solvable. Its base subgroup ${⨁}_{\mathbb{Z}}{C}_2$ is not finitely generated. Every subgroup of a polycyclic group is finitely generated, so the lamplighter group is not polycyclic.

Let $N$ be the subgroup whose upper-left block is $I_2$. It is a finitely generated nilpotent subgroup of the integer upper unitriangular group $U{T}_4(\mathbb{Z})$ and is normal in $L$. The block-diagonal matrixgenerates an infinite cyclic quotient, soFinitely generated nilpotent groups are polycyclic, and an extension of polycyclic groups is polycyclic. Hence $L$ is polycyclic.

Inside $L$, retain only $t$ and the entries in positions $(1,3)$ and $(2,3)$. They form a subgroupThe characteristic polynomial of $A$ is ${\lambda }^2-5\lambda +1$, so its eigenvalues areOne has modulus greater than one, and the resulting semidirect product has exponential growth. Every finitely generated virtually nilpotent group has polynomial growth, as does each of its finitely generated subgroups. Therefore $L$ cannot be virtually nilpotent.

The empty word is the unique vertex of degree three in the underlying tree; every other vertex has degree four. Every graph automorphism therefore fixes the empty word and permutes its three neighbours. Those neighbours are precisely $\{0,1,2\}$, so this set is invariant.

The automorphism $a$ cyclically permutes the first letter and leaves the remaining suffix unchanged, so $a^3=1$ and $a\ne 1$.

In the section notation,Thus $b^3=(a^3,1,b^3)=(1,1,b^3)$. An automorphism $c$ satisfying $c=(1,1,c)$ fixes every finite word: repeatedly entering the third subtree eventually reaches the end of the word. Hence $b^3=1$. Since $b(00)=01$, it is nonidentity, and both $a$ and $b$ have order three.

The action on the first level defines a surjective homomorphismthat sends $a$ to the displayed cycle and $b$ to the identity. Its kernel is therefore the normal closure of $b$, proving that $\{b\}$ normally generates ${\mathrm{Stab}}_G(1)$.

Apply the Reidemeister–Schreier theorem with transversal $\{1,a,a^{-1}\}$. The generators arising from $a$ are trivial, while those arising from $b$ areConsequently these three elements generate ${\mathrm{Stab}}_G(1)$.

An element $g$ fixing the first level restricts to an automorphism $g_i$ on each rooted subtree, and composition is coordinatewise. Thereforeis a homomorphism. If all three sections are trivial, $g$ fixes every word, so $\varphi$ is injective.

Directly from the recursions,The generators found in the preceding part therefore have every section in $G$, so $\varphi ({\mathrm{Stab}}_G(1))\subseteq G^3$.

Moreover, every coordinate projection of this image contains both $a$ and $b$, and is therefore onto $G$. Since $a$ acts transitively on the first level, induction shows that $G$ acts transitively on every level of the rooted tree. The $n$th level has $3^n$ vertices, so the orders of these finite orbits are unbounded. Hence $G$ is infinite.

Use the convention $[g,h]=g^{-1}{h}^{-1}gh$. In $G/K$ the images of $a$ and $b$ commute and both have order three, so $G/K$ is a quotient of $C_3\times C_3$. Thus

Put $d=a^{-1}ba$. From the preceding section calculations,Since both elements fix the first level, their commutator is computed coordinatewise, andThe element $[d,x]$ belongs to $K$. The third-coordinate projection of $\varphi ({\mathrm{Stab}}_G(1))$ is onto $G$, so conjugating this element inside the stabilizer shows that $\varphi (K)$ contains $(1,1,x^g)$ for every $g\in G$. Because the conjugates $x^g$ generate $K$, it contains $1\times 1\times K$. Conjugation by $a$ cyclically permutes the coordinates; hence it also contains $K\times 1\times 1$ and $1\times K\times 1$. These coordinate subgroups commute, givingInjectivity of $\varphi$ identifies its inverse image with a subgroup of $K$ isomorphic to $K^3$.

A group $G$ is residually finite when, for every $1\ne g\in G$, there are a finite group $Q$ and a group homomorphism $q:G\to Q$ such that $q(g)\ne 1$. Equivalently, the intersection of all finite-index normal subgroups of $G$ is trivial.

By the Fundamental theorem of finitely generated abelian groups,with $T$ finite. If a nonzero element has a nonzero component in $T$, projection to $T$ separates it. Otherwise some integer coordinate is a nonzero $n$; choose a prime $p$ not dividing $n$ and reduce that coordinate modulo $p$. This gives a finite quotient in which the element survives, so every finitely generated abelian group is residually finite.

If $G$ is generated by $d$ elements, a homomorphism $G\to Q$ is determined by the images of those generators. There are at most $|Q{|}^d$ such choices, so only finitely many homomorphisms exist.

Let $\alpha :G\to G$ be surjective and suppose that $1\ne g\in \ker \alpha$. By residual finiteness, choose $q:G\to Q$ with $Q$ finite and $q(g)\ne 1$. The preceding part makes the sequencerepeat, so $q{\alpha }^i=q{\alpha }^j$ for some $i<j$. Surjectivity of ${\alpha }^i$ permits cancellation on the right and gives $q=q{\alpha }^{j-i}$. But $g\in \ker {\alpha }^{j-i}$, which would imply $q(g)=1$, a contradiction. Thus $\alpha$ is injective. Every finitely generated residually finite group is therefore a Hopfian group.

Enumerate $X=\{x_1,\dots ,x_n\}$ and $Y=\{y_1,\dots ,y_n\}$. The universal property of a free group gives an endomorphismSince $Y$ generates, $\alpha$ is surjective. The finitely generated free group is residually finite and hence Hopfian by the preceding part, so $\alpha$ is an automorphism. An automorphism sends a free basis to a free basis; therefore $Y$ is a basis of $F(X)$.

For the free basis $X=\{x_0,x_1,x_2,\dots \}$, defineThe universal property of a free group extends this assignment to an endomorphism of $F(X)$. It is surjective because every $x_n$ is the image of $x_{n+1}$, but it is not injective because $x_0\ne 1$ lies in its kernel. Thus $F(\mathbb{N})$ is not Hopfian.

Letbe the group of finitely supported functions $f:G\to H$ with pointwise multiplication. The left-translation actiondefines the restricted wreath product

Suppose $X$ and $Y$ are finite generating sets for $G$ and $H$. Embed each $y\in Y$ as a lamp supported at the identity of $G$. Conjugating these lamps by words in $X$ produces copies of $Y$ at every coordinate, and these copies generate $B$. Thus $X$ together with the identity-coordinate copy of $Y$ is a finite generating set for $H\wr G$.

Let $\theta :H\wr G\to Q$ be any homomorphism to a finite group. Because $G$ is infinite, two distinct elements $r,s\in G$ have $\theta (r)=\theta (s)$. For $h\in H$, denote by $h_g$ the lamp with value $h$ at $g$. Conjugation translates lamps, sofor every $h\in H$. Choose $h,k\in H$ with $[h,k]\ne 1$. Lamps at different coordinates commute, and thereforeBut $[h_r,k_r]$ is the nonidentity lamp $[h,k{]}_r$. This same nonidentity element is killed by every finite quotient, so $H\wr G$ is not residually finite.

Choose $1\ne h\in H$ and let $t$ generate $\mathbb{Z}$. For each binary string $({\epsilon }_0,\dots ,{\epsilon }_{n-1})$, the wordrecords that string in the lamps at positions $0,1,\dots ,n-1$. The resulting $2^n$ group elements are distinct and have word length at most $3n$ with respect to any finite generating set containing $h$ and $t$. Hence the growth function is bounded below exponentially, and $H\wr \mathbb{Z}$ has exponential growth.

Write an element of $H\wr \mathbb{Z}$ as $(f,t^n)$. If it is central, commuting with $t$ makes the finitely supported lamp configuration $f$ invariant under translation. The only such configuration on the infinite set $\mathbb{Z}$ is the identity, so $f=1$. If $n\ne 0$, then $t^n$ moves a nonidentity lamp at position zero to position $n$ and does not commute with it. Therefore $n=0$, and