= Odd-prime principal p-adic congruence subgroup is uniform
{title2=$U_m=1+p^mM_d(\mathbb Z_p),\quad[U_m:U_{m+1}]=p^{d^2}$}
For an odd prime $p$, the <matrix logarithm> and <matrix exponential> are inverse maps between $1+pM_d(\mathbb Z_p)$ and $pM_d(\mathbb Z_p)$. They imply that the $p$th-power map carries $U_m$ bijectively onto $U_{m+1}$. Direct multiplication gives $[U_i,U_j]\subseteq U_{i+j}$, so the <group> is a <powerful pro-p group>. The quotient $U_m/U_{m+1}$ is the additive <matrix> space over $\mathbb F_p$, with constant index $p^{d^2}$. The $d^2$ <matrices> $1+pE_{ij}$ topologically generate, by successive approximation through these quotients. The <lower p-series> is exactly $U_m$, so the <group> is a <uniform pro-p group> and a <torsion-free group>. At $p=2$ the first congruence subgroup contains $-I$ and is not uniform.
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