Past exam of the short options course of the University of Oxford / 2023 / 2 (Group theory)

A group presentation $⟨X\mid R⟩$ denotes the quotient $F(X)/⟨⟨R⟩⟩$ of the free group on the generator symbols by the normal closure of the relator words. The symbols generate the quotient, and each relator becomes the identity. The phrase also asserts that no additional relations are imposed except consequences of these relators and the group axioms. Equivalently, an assignment of the generators to a group extends to a group homomorphism from the presented group exactly when all relators evaluate to the identity; this is the universal property of a group presentation.

Multiplication, followed by induction and inversion for negative powers, givesIf $c\ne 0$, apply Euclidean division to write $a=q|c|+r$ with $0\le r<|c|$. Choose $n=-q\mathrm{sgn}(c)$, so $a+nc=r$. This is one step of the Euclidean algorithm implemented by a determinant-one row operation.

After the reduction in part (b), multiply by $S$. The new bottom-left entry is $r$, whose absolute value is strictly smaller than the previous $|c|$:Repeated Euclidean division must terminate because the nonnegative integers $|c|$ decrease strictly until zero. Thus a word $W$ in $S,T$ and their inverses gives a matrix with bottom-left entry zero. Its determinant is one, so the diagonal entries are equal and are either both $1$ or both $-1$:Since $S^2=-I$, this final matrix belongs to $⟨S,T⟩$. So does $W^{-1}$, proving

The identity $T=S^{-1}(ST)$ shows that $S,ST$ generate the same group. Direct multiplication givesConsequently their images $\bar{S},\overline{ST}$ in the quotient group $PSL(2,\mathbb{Z})$ satisfy ${\bar{S}}^2=1$ and ${\overline{ST}}^3=1$. The universal property of a group presentation gives a group homomorphism taking $u\mapsto \bar{S}$, $v\mapsto \overline{ST}$ from the stated presentation. It is surjective because these images generate the target. Injectivity is not needed here.

Suppose $N$ were a normal subgroup of index $m$ and put $Q=SL(2,{\mathbb{F}}_p)/N$. The supplied reduction map and part (c) show that $Q$ is generated by the images $s,t$ of $S,ST$. Their orders divide $4$ and $6$, respectively, because $S^4=I$ and $(ST{)}^6=I$. By Lagrange's theorem, their orders also divide $|Q|=m$. Since $\gcd (m,6)=1$, both $\gcd (m,4)$ and $\gcd (m,6)$ equal one. Thus $s=t=1$, forcing $Q$ to be trivial and $m=1$, a contradiction. No such normal subgroup exists for $m>1$.