Block idempotents lift uniquely from kH to OH

Put $A=\mathcal{O}H$. It is finite free over $\mathcal{O}$, hence complete and separated for the $\mathfrak{m}$-adic topology by F1. Every element of $\mathfrak{m}A$ lies in $J(A)$: if $x\in \mathfrak{m}A$ and $y\in A$, then $(yx{)}^n\to 0$, so $1-yx$ has inverse ${\sum }_{n\ge 0}(yx{)}^n$. This is the Jacobson-radical test.

Let $\bar{e}\in A\mathrm{/}\mathfrak{m}A=kH$ be idempotent and choose any lift $e_1\in A$. Inductively, if $e_n^2-e_n\in {\mathfrak{m}}^{n}A$, apply the Newton correction $e_{n+1}=e_n-(2e_n-1)(e_n^2-e_n)\mathrm{.}$ Because $e_n$ commutes with $e_n^2-e_n$, direct expansion shows that $e_{n+1}^2-e_{n+1}$ is a multiple of $(e_n^2-e_n{)}^2$, while $e_{n+1}-e_n\in {\mathfrak{m}}^{n}A$. Thus the errors tend to zero and $(e_n)$ is Cauchy. Completeness gives a limit $e$ with $e^2=e$ and reduction $\bar{e}$.

Suppose now that $\bar{e}$ is central. For $X=eA(1-e)$ its reduction is $\bar{e}(kH)(1-\bar{e})=0$, hence $X=\mathfrak{m}X$: the nontrivial inclusion uses that $\mathfrak{m}=(\pi )$ in the DVR, since $x=\pi a=e(\pi a)(1-e)=\pi (ea(1-e))$. The finite $\mathcal{O}$-module $X$ has generators $x_1,\dots ,x_t$ with $x_i=\pi {\sum }_j{a}_{ij}{x}_j$. Multiplying $(I-\pi (a_{ij}))(x_j)=0$ by its adjugate shows that $\det (I-\pi (a_{ij}))x_j=0$ for every $j$. The determinant is congruent to $1$ modulo $\mathfrak{m}$, hence is a unit, so $X=0$. Applying the same argument to $(1-e)Ae$ gives that space zero too. Therefore $ea=eae=ae$ for every $a\in A$, and $e$ is central.

If central idempotents $e,f\in A$ have the same reduction, then $e-f\in \mathfrak{m}A\subseteq J(A)$. The commuting products $e(1-f)$ and $f(1-e)$ are idempotents in $J(A)$, and an idempotent in the Jacobson radical is zero. Hence $e=ef=f$. Thus every central idempotent of $kH$ has exactly one central lift.

Central primitivity is preserved. If a central lift $e$ decomposed into two nonzero orthogonal central idempotents, neither summand could reduce to zero, since step 1.1 puts its kernel inside $J(A)$; their reductions would decompose $\bar{e}$. Conversely, a central decomposition of $\bar{e}$ lifts termwise by steps 2.1–3.1, and uniqueness in step 4.1 makes the lifted sum equal to $e$. Hence $e$ is primitive exactly when $\bar{e}$ is primitive. Together with F2 this proves the bijection and the asserted unique block lift, including the trivial-group case. The constructions are finite or sequential limits fixed by explicit formulas, so no choice principle is used.