Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 104 / 3

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 $K^3$.