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A filtration on a group is a function $\omega :G\to \mathbb{R}\cup \{\infty \}$ satisfyingIt is a p-valuation when it is separated and, for $x\ne 1$,

Let $p$ be odd and letFor $g=1+A\ne 1$, define $\omega (g)={\min }_{i,j}{v}_p(A_{ij})$. Matrix multiplication and the identitygive the two filtration inequalities. Since $\omega (g)\ge 1>1/(p-1)$, only the p-power condition remains. The binomial theorem givesThe first term has valuation $1+\omega (g)$, while every other term has strictly larger valuation because $p$ is odd and $\omega (g)\ge 1$. Hence $\omega (g^p)=\omega (g)+1$, so this is a p-valuation.

Finally, if ${\omega }_1,{\omega }_2$ are p-valuations, put $\omega =\min ({\omega }_1,{\omega }_2)$. Taking minima preserves both filtration inequalities and the strict lower bound, whileIf $\omega (g)=\infty$, both original valuations force $g=1$. Therefore the pointwise minimum of two p-valuations is again a p-valuation.

For a separated filtered group defineThe associated graded Lie algebra of a filtered group isEach quotient is abelian because $[G_{\lambda },G_{\lambda }]\subseteq G_{2\lambda }\subseteq G_{\lambda +}$. If $x\in G_{\lambda }$ and $y\in G_{\mu }$, setThe standard commutator identities $[xy,z]=[x,z{]}^y[y,z]$ and $[x,yz]=[x,z][x,y{]}^z$ prove well-defined bilinearity, while the Hall-Witt identity gives the Jacobi identity. If $\omega ([x,y])=\omega (x)+\omega (y)$, this bracket has nonzero initial form, so the graded Lie algebra is nonabelian.

For a p-valuation, defineThe p-power axiom and the Hall-Petrescu formula make this well defined and turn each homogeneous component into part of a graded ${\mathbb{F}}_p[t]$-module. The leading-term congruence $[x^p,y]\equiv [x,y{]}^p$ modulo terms of valuation greater than $\omega (x)+\omega (y)+1$ makes the bracket ${\mathbb{F}}_p[t]$-bilinear.

For the given upper-triangular group, write an element of degree $n$ to leading order asLet $X_n,Y_n$ denote the classes with $(\alpha ,\beta )=(1,0),(0,1)$. Matrix commutators giveand pth powers give $t{X}_n=X_{n+1}$ and $t{Y}_n=Y_{n+1}$. Thus