= Solution
A <filtration on a group> is a function $\omega:G\to\mathbb R\cup\{\infty\}$ satisfying
$$
\omega(xy^{-1})\geq\min\{\omega(x),\omega(y)\},
\qquad
\omega([x,y])\geq\omega(x)+\omega(y).
$$
It is a <p-valuation> when it is separated and, for $x\ne1$,
$$
\omega(x)>\frac1{p-1},
\qquad
\omega(x^p)=\omega(x)+1.
$$
Let $p$ be odd and let
$$
G=\ker\bigl(\operatorname{GL}_2(\mathbb Z_p)\to\operatorname{GL}_2(\mathbb F_p)\bigr)
=1+pM_2(\mathbb Z_p).
$$
For $g=1+A\ne1$, define $\omega(g)=\min_{i,j}v_p(A_{ij})$. Matrix multiplication and the identity
$$
[1+A,1+B]-1=(1+A)^{-1}(1+B)^{-1}(AB-BA)
$$
give the two filtration inequalities. Since $\omega(g)\geq1>1/(p-1)$, only the p-power condition remains. The <binomial theorem> gives
$$
(1+A)^p-1=pA+\sum_{i=2}^{p-1}\binom piA^i+A^p.
$$
The first term has valuation $1+\omega(g)$, while every other term has strictly larger valuation because $p$ is odd and $\omega(g)\geq1$. 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, while
$$
\omega(g^p)=\min_i\bigl(\omega_i(g)+1\bigr)=\omega(g)+1.
$$
If $\omega(g)=\infty$, both original valuations force $g=1$. Therefore \b[the pointwise minimum of two p-valuations is again a p-valuation].
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