 Solution

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.