= Solution
The reduction map sends $H_k$ onto $H_{k-1}$ with kernel of order $p$, so $|H_k|=p^{k-1}$. More explicitly, the lifting-the-exponent calculation
$$
v_p((1+p)^{p^j}-1)=j+1
$$
shows that $1+p$ has order $p^{k-1}$ modulo $p^k$. Hence $H_k=\langle1+p\rangle$ is cyclic of that order.
For the <matrix> claim, suppose $A\ne I$ and choose $m$ maximal such that $A=I+p^mB$, so some entry of $B$ is nonzero modulo $p$. Replacing $A$ by a suitable power, we may assume its order is a prime $q$. If $q\ne p$, binomial expansion modulo $p^{m+1}$ gives
$$
A^q\equiv I+qp^mB\not\equiv I\pmod {p^{m+1}}.
$$
If $q=p$, expansion modulo $p^{m+2}$ gives
$$
A^p\equiv I+p^{m+1}B\not\equiv I\pmod {p^{m+2}},
$$
because $p$ is odd. Both contradict $A^q=I$, so $A=I$.
For $p=2$, $r=1$ fails because $-I=I+2(-I)$ has order two. If $r=2$, the same argument works; in the order-two case, for maximal $m\geq2$,
$$
(I+2^mB)^2=I+2^{m+1}(B+2^{m-1}B^2)
e I,
$$
since the parenthesis is nonzero modulo two. Thus the smallest value is $r=2$.
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