Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 134 / 3

A group $G$ is residually finite if, for every $1\ne g\in G$, there are a finite group $Q$ and a homomorphism $\varphi :G\to Q$ such that $\varphi (g)\ne 1$.

Let $F(S)$ be a free group and let $1\ne w\in F(S)$ be a reduced word. In the rose with one oriented loop for each $s\in S$, the word $w$ determines a nonclosed reduced path from a chosen vertex in the universal covering tree. Its finite image path can be completed to a finite covering graph of the rose: for each label, pair the still unmatched incoming and outgoing edge germs, adding finitely many vertices if necessary. The lift of $w$ remains nonclosed in this finite cover.

Let $H\le F(S)$ be the finite-index subgroup represented by this based cover. Then $w\notin H$. The action of $F(S)$ on the finite set of right cosets $H\backslash F(S)$ gives a homomorphism to a finite symmetric group, and $w$ does not fix the coset $H$. Thus $w$ survives in a finite quotient. Since $w$ was arbitrary, every free group is residually finite.

Let $X$ be a subcomplex of a product of graphs $A\times B$. Orient every edge of each factor. An edge of $X$ is horizontal or vertical according to its factor, and this type is preserved across opposite sides of every square.

A hyperplane of a cube complex of horizontal type retains one fixed edge of $A$ while moving through edges of $B$; the analogous statement holds vertically. The factor orientation makes every hyperplane two-sided. A square has one horizontal and one vertical direction, so no hyperplane self-intersects. At a vertex there is at most one incident edge with a fixed factor edge and orientation, so no hyperplane self-osculates. Finally, a horizontal hyperplane and a vertical hyperplane can cross only in the unique product square determined by their two factor edges. If that square belongs to $X$, it fills every corner at which those two dual edges meet; if it does not, the hyperplanes never cross. Thus no pair interosculates. All four hyperplane pathologies are absent, so $X$ is a special cube complex.

Call the two horizontal edge classes indicated by one and two arrowheads $a$ and $b$. All vertices in the displayed quotient are identified. In the third displayed square, an $a$-edge and a $b$-edge are opposite, so they are dual to the same hyperplane of a cube complex $H$. At the unique vertex, the distinct edges $a$ and $b$ are therefore dual to $H$, but no square has them as adjacent sides. With the orientations shown, they have the same initial vertex. Hence $H$ is a self-osculating hyperplane, one of the forbidden pathologies of a special cube complex. The displayed cube complex is not special.

Because $X$ is special, the fundamental group of a special cube complex embeds in a Right-angled Artin group. Right-angled Artin groups are residually finite by the residual finiteness of a right-angled Artin group, and a subgroup of a residually finite group is residually finite. Hence ${\pi }_{1}X$ is residually finite.

If ${\pi }_{1}X$ were simple, choose $1\ne g\in {\pi }_{1}X$. A finite quotient in which $g$ survives has a proper normal kernel. Simplicity would force that kernel to be trivial, embedding ${\pi }_{1}X$ into a finite group, contrary to the assumption that ${\pi }_{1}X$ is infinite. This is precisely the obstruction that an infinite residually finite group is not simple.