Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 145 / 3 / Solution

Give the generator $p$ weight one and $g-1$ weight $\omega (g)$. For $\lambda \ge 0$, let ${\mathbb{Z}}_p[G{]}_{\lambda }$ be the ${\mathbb{Z}}_p$-span ofThe identitiesandtogether with the filtration inequalities show that the definition is independent of how an element is expanded and thatCompleteness and an Ordered basis of a complete p-valued group give separatedness through unique noncommutative power-series expansions. This is the Lazard filtration on a group algebra.

Multiplication by $p$ induces a central degree-one element $t$, so $\mathrm{gr}{\mathbb{Z}}_p[G]$ is a graded ${\mathbb{F}}_p[t]$-algebra. Sending the initial form of $g$ to the initial form of $g-1$ respects the Lie bracket, since the displayed algebra commutator has leading term represented by $[g,h]-1$. The universal property of the universal enveloping algebra therefore gives the Lazard enveloping-algebra mapIt is surjective because the defining filtered pieces are spanned by products of $p$ and the elements $g-1$.

For the principal congruence subgroup of ${\mathrm{SL}}_2({\mathbb{Z}}_p)$, let $e=E-1$, $f=F-1$, and $h=H-1$. In the group algebra,The factor $FE$ is congruent to one in filtration degree zero. Direct matrix multiplication, now applied to the group commutator, givesConsequently $ef-fe\equiv H^p-1\equiv p(H-1)=ph$ modulo ${\mathbb{Z}}_p[G{]}_3$. The same argument starts fromUsing $(1+p{)}^{-1}=1-p+p^2+O(p^3)$ in the two matrix commutators givesSince the prefactors $EH,FH$ may again be replaced by one at this precision, we obtain
Therefore