 Solution

The first alternative in the supplied analytic-group theorem would give a faithful representation of $\mathcal{N}$ in some ${\mathrm{GL}}_d({\mathbb{Z}}_p)$ or ${\mathrm{GL}}_d({\mathbb{F}}_p[[t]])$. 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 $\phi (p^a)=p^{a-1}(p-1)$ 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 $p^e\ge d$,Thus its order is at most $p^{\lceil {\log }_{p}d\rceil }$, again a fixed bound. Therefore the linear alternative is impossible in both coefficient rings.