Past exam of the mathematics course of the University of Cambridge / 2013 / ia / Paper 3 / 6D (Groups)

For $M=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ in the special linear group over a finite field, define the Möbius transformation $M\cdot x=(ax+b)/(cx+d)$ when the denominator is nonzero, and set $M\cdot x=\infty$ when it is zero. At the extra point, $M\cdot \infty =a/c$ for $c\ne 0$, and $M\cdot \infty =\infty$ for $c=0$. Determinant one prevents simultaneous zero numerator and denominator.

The orbit-stabiliser theorem states $|G\cdot x|=[G:G_x]$, and for a finite group $|G|=|G\cdot x||G_x|$. The matrix $\left(\begin{array}{cc}t& -1\\ 1& 0\end{array}\right)$ sends infinity to any chosen $t\in {\mathbb{F}}_p$, so the orbit of infinity is the whole projective line, of size $p+1$. Its stabilizer subgroup iswhich has $p(p-1)$ elements. ConsequentlyThe SL2 action on a finite projective line need not be faithful: scalar matrices can fix every projective point. Faithfulness is not needed for this orbit calculation.

For $U_t=\left(\begin{array}{cc}1& t\\ 0& 1\end{array}\right)$ and $D_r=\mathrm{diag}(r,r^{-1})$, direct multiplication gives $D_r{U}_1{D}_r^{-1}=U_{r^2}$. In ${\mathbb{F}}_{11}$, $5^2=3$ and $5^{-1}=9$, soTo rule out all conjugators when $p=5$, suppose $PA{P}^{-1}=B$ and write $P=\left(\begin{array}{cc}r& s\\ t& u\end{array}\right)$. Comparing $PA=BP$ gives $t=0$ and $r=3u$. Its determinant condition is $ru=1$, so $r^2=3$. The nonzero squares modulo five are $1$ and $4$, and therefore no such $P$ exists. The two matrices are not conjugate in ${\mathrm{SL}}_2(5)$. This is the square-class obstruction in unipotent conjugacy in SL2 over a finite field; checking only diagonal conjugators would not by itself exclude every possible conjugator.

Write the matrix as $M(a,b,x)$. Direct multiplication and inversion giveThe identity is $M(0,0,0)$ and all determinants are one, so these formulas verify the subgroup criterion inside ${\mathrm{GL}}_3(\mathbb{R})$. This is the real Heisenberg group.

The map $M(a,b,x)\mapsto a$ is a surjective group homomorphism to the additive group of real numbers; its kernel is exactly $H$. Hence $H$ is normal and the first isomorphism theorem givesTwo matrices commute exactly when $a{b}^{\prime }=a^{\prime }b$. Requiring that relation for every $a^{\prime },b^{\prime }$ forces $a=b=0$, while $x$ is unrestricted. ThereforeFinally $M(a,b,x)\mapsto (a,b)$ is a surjective group homomorphism to the additive group ${\mathbb{R}}^2$ with precisely this kernel of a group homomorphism, giving $G/Z(G)\cong ({\mathbb{R}}^2,+)$. The central coordinate records noncommutativity: the group commutator is $M(0,0,a{b}^{\prime }-a^{\prime }b)$.