= Solution
The first alternative in the supplied analytic-group theorem would give a <faithful representation> of $\mathcal N$ in some $\operatorname{GL}_d(\mathbb Z_p)$ or $\operatorname{GL}_d(\mathbb F_p[[t]])$. The embedding proved above supplies cyclic <subgroups> of order $p^a$ for arbitrarily large $a$. Such orders are impossible in either fixed <matrix> <dimension>.
In <characteristic zero>, a <matrix> $A$ of order $p^a$ is diagonalizable over an algebraic <closure>, because its <minimal polynomial> divides the <separable polynomial> $X^{p^a}-1$. At least one <eigenvalue> must be a primitive $p^a$th <root of unity>; otherwise $A^{p^{a-1}}=I$. The <cyclotomic polynomial> $\Phi_{p^a}$ is irreducible over $\mathbb Q_p$, since $\Phi_{p^a}(1+X)$ satisfies the <Eisenstein criterion> at $p$. Consequently that <eigenvalue> has degree $\varphi(p^a)=p^{a-1}(p-1)$ over $\mathbb Q_p$, which must be at most the <matrix> <dimension> $d$. This bounds $a$.
In characteristic $p$, pass from the power-series ring to its fraction <field>. A <matrix> of $p$-power order satisfies $X^{p^a}-1=(X-1)^{p^a}$, so it is unipotent. Writing $A=I+B$, nilpotence gives $B^d=0$. For $p^e\ge d$,
$$
A^{p^e}=I+B^{p^e}=I.
$$
Thus its order is at most $p^{\lceil\log_pd\rceil}$, again a fixed bound. Therefore \b[the linear alternative is impossible] in both coefficient rings.
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