Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 4 / 5

For a group presentation $\mathcal{P}=⟨X\mid R⟩$ with $X$ finite, let $F(X)$ be its free group. For a nontrivial relator defineThe p-deficiency in the unshifted convention used here isIf the weighted sum diverges the value is $-\infty$; identity relators may be omitted or assigned weight zero. Roots are taken in the free group, not in the presented quotient. Some authors subtract one from this definition; here the requested threshold is $1$.

Two elementary bounds explain why p-deficiency detects infinitude. The p-rank of a group isHere $G^p$ denotes the subgroup generated by all $p$th powers. Each relator that is not a $p$th power imposes at most one linear relation in this vector space, and a $p$th-power relator imposes none. If there are $s$ relators of the first type, then

The second bound is the index-p rewriting bound for p-deficiency. Suppose $H⊴G$ has index $p$, and its preimage in $F(X)$ is $V$. The Nielsen–Schreier formula gives $V$ rank $1+p(|X|-1)$. For a relator $r=w^{p^k}$, there are two cases in the Reidemeister–Schreier theorem. If $w\in V$, its $p$ coset-conjugates are all $p^k$th powers in $V$, with total weight at most $p{p}^{-k}$. If $w\notin V$, then $k\ge 1$, since $r\in V$. Its cosets generate $F(X)/V$, so representatives $1,w,\dots ,w^{p-1}$ show that the $p$ rewritten conjugates of $r$ are redundant up to conjugation in $V$. One relator suffices, and $r=(w^p{)}^{p^{k-1}}$ has weight at most $p^{-(k-1)}$. In both cases the total weight is at most $p$ times the old weight. Thus the induced group presentation $\mathcal{Q}$ of $H$ satisfiesThe argument applies termwise to infinitely many relators whenever the weighted sum converges.

If ${\mathrm{def}}_p(\mathcal{P})\ge 1$, the p-rank of a group bound gives a surjection to $C_p$, hence a normal subgroup of index $p$. The rewriting bound gives that subgroup another presentation of p-deficiency at least one. Iterating produces subgroups of index $p^j$ for every $j$. p-deficiency at least one implies infinitude.

Now enumerate the nonidentity elements $w_1,w_2,\dots$ of $F(x,y)$ and choose the presentationIts p-deficiency obeysThe infinitude criterion shows that $T$ is infinite. It is generated by two elements, and every element is represented by some $w_i$ or is the identity; the imposed relation makes its order a power of $p$. Thus $\boxed{T\text{ is an infinite finitely generated torsion group}}$. This is a torsion group construction by p-power relators; the presentation intentionally has infinitely many relators.

No prime and no presentation of $G_1$ have p-deficiency at least one. Abelianizing the cyclic squaring relations makes each generator zero: for example $vw{v}^{-1}=w^2$ becomes $w=2w$, so $w=0$, and the other four relations kill $x,y,z,v$. Thus the abelianization of $G_1$ is trivial and $d_p(G_1)=0$ for every prime number $p$. The presentation-independent p-rank of a group bound from the general solution givesfor every group presentation $\mathcal{P}$ of $G_1$. This rules out alternative presentations, not just the one displayed.

Yes, for $p=2$. The displayed group presentation is that of the infinite dihedral group. Put $u=at$ and $v=t$. Then $v^2=1$ andConversely, from $u^2=v^2=1$, set $a=uv$ and $t=v$; then $ta{t}^{-1}=vu=a^{-1}$. These inverse substitutions giveEach relator has free-group $2$-root exponent one, soThe change of presentation matters: the original presentation's mixed conjugation relator has weight one and would give a smaller p-deficiency.

No prime and no presentation of $G_3$ have p-deficiency at least one. Set $r=xy$. The relations giveEvery element has form $r^j$ or $r^{j}x$ with $0\le j<1024$. Conversely, the usual rotations and reflections of a regular $1024$-gon satisfy the presentation and give $2048$ distinct elements. Hence $G_3$ is the finite dihedral group of order $2048$. The criterion p-deficiency at least one implies infinitude excludes every alternative presentation and every prime. As a check, the given presentation has

Yes: the displayed presentation already has p-deficiency exactly one for $p=2$. The words $x,y,z,xy,yz,zx$ are not proper powers in the free group. For the length-two words this follows directly from their distinct consecutive letters in a cyclically reduced word. Thus each relator's $2$-root exponent is precisely the exponent of $2$ in its displayed power. The relator weights, in the given order, areTheir sum is $2$, givingIn particular the infinitude criterion proves that this group is infinite, although the question only asks for the existence of the presentation and prime.