Fix a prime $p$ and let $G_n=\operatorname{Gal}(\mathbb{Q}(\zeta_{p^{n+1}})/\mathbb{Q})$. We can identify $G_n$ with $(\mathbb{Z}/p^{n+1}\mathbb{Z})^\times$ by mapping $a$ mod $p^{n+1}$ to $\sigma_a:\zeta_{p^{n+1}} \mapsto\zeta_{p^{n+1}}^a$. There is a (unique, if $p$ is odd) decomposition $G_n=\Delta\times \Gamma_n$, where $\Delta\cong (\mathbb{Z}/p\mathbb{Z})^\times$ and $\Gamma_n\cong\mathbb{Z}/p^n\mathbb{Z}$. One can now define an 'obvious' map $\mathbb{Z}_p[G_{n}]\rightarrow \mathbb{Z}_p[\Gamma_n]$ as follows. Let $\omega :(\mathbb{Z}/p\mathbb{Z})^\times\rightarrow \mathbb{Z}_p^\times$ be the mod $p$ cyclotomic character and $\langle a \rangle$ the projection of $a \in (\mathbb{Z}/p^{n+1}\mathbb{Z})^\times$ onto $\mathbb{Z}/p^n\mathbb{Z}$, so that $\sigma_{\langle a \rangle}\in \Gamma_n$. We then have a map $$ \omega^i:\mathbb{Z}_p[G_n]\rightarrow \mathbb{Z}_p[\Gamma_n] $$ given by $\sum_{a \in (\mathbb{Z}/p^{n+1}\mathbb{Z})^\times}c_a\sigma_a\mapsto \sum_{a \in (\mathbb{Z}/p^{n+1}\mathbb{Z})^\times}\omega^i(a)c_a\sigma_{\langle a \rangle}$, where $c_a\in \mathbb{Z}_p$ and we are extending $\omega$ to $(\mathbb{Z}/p^{n+1}\mathbb{Z})^{\times}$ by precomposing with reduction mod $p$.
There is also another way to define an element of $\mathbb{Z}_p[\Gamma_n]$ from $\mathbb{Z}_p[G_n]$. We can view $\mathbb{Z}_p[G_n]$ as a $\mathbb{Z}_p[\Delta]$-module and take the decomposition coming from orthogonal idempotents. More specifically, if $$ \epsilon_i=\frac{1}{p-1}\sum_{\sigma\in \Delta}\omega^i(\tau)\tau^{-1} $$ and if $\vartheta\in \mathbb{Z}_p[G_n]$ we get that $\epsilon_i\vartheta$ lies in the $\omega^i$-eigenspace of $\mathbb{Z}_p[G_n]$, i.e, a copy of $\mathbb{Z}_p[\Gamma_n]$ where $\tau\in \Delta$ acts by multiplication by $\omega^i(\tau)$.
My intuition tells me that these two methods of defining an element of $\mathbb{Z}_p[\Gamma_n]$ by 'projection' from $\mathbb{Z}_p[G_n]$ should be the same, i.e., that \begin{equation}\tag{1} \epsilon_i\vartheta=\omega^i(\vartheta), \end{equation} but for some reason I'm having a hard time showing this. Is this equation (1) true?
I think this is either easy and I'm getting myself confused or possibly it's just not true... In any case, here are some thoughts: In general, one has $\tau\epsilon_{\chi}=\chi(\tau)\epsilon_{\chi}$ for a character $\chi\in \hat \Delta$. Thus, since we can decompose $\sigma_a=\tau_a\sigma_{\langle a\rangle}$ for some $\tau_a\in \Delta$, we have $$ \epsilon_i \vartheta= \epsilon_i \sum_{a\in (\mathbb{Z}/p^{n+1}\mathbb{Z})^\times}\omega^i(\tau_a)c_a \sigma_{\langle a \rangle}=\frac{1}{p-1}\sum_{\tau\in \Delta}\sum_{a\in (\mathbb{Z}/p^{n+1}\mathbb{Z})^\times}\omega^i(\tau\tau_a)c_a \sigma_{\langle a \rangle}\tau^{-1}. $$ After staring at this for far too long and fiddling with reindexing, I am still not sure why this should be equal to $\omega^i(\vartheta)=\sum_{a \in (\mathbb{Z}/p^{n+1}\mathbb{Z})^\times}\omega^i(a)c_a\sigma_{\langle a \rangle}$...
For $p$ odd prime with $g$ of order $p-1$ and $1+p$ of order $p^{n-1}$ the generators of $G_n=\Bbb{Z/p^nZ}^\times$. Let $r(p-1)=1\bmod p^{n-1}$ so that $\sigma\to \sigma^{r(p-1)}$ is the natural projection $G_n\to G_n^{p-1}=\Gamma_n$.
Let $$e_j = \frac1{p-1}\sum_{k=1}^{p-1} g^k \zeta_{p-1}^{kj}\in \Bbb{Z}_p[G_n], \qquad e_j^2 = e_j, \qquad \omega(g^m (1+p)^r) = \zeta_{p-1}^m\in \Bbb{Z}_p$$
Then for $\sigma\in G_n$ $$e_j \sigma =e_j\sigma^{r(p-1)}\omega(\sigma)^{-j} \in \Bbb{Z}_p[G_n]$$
The map (non-unital ring endomorphism of $\Bbb{Z}_p[G_n]$) $$\sigma \to e_j \sigma^{r(p-1)}\omega(\sigma)^{-j} $$ factors through the ring homomorphism $$\sigma\to \sigma^{r(p-1)}\omega(\sigma)^{-j} ,\qquad \Bbb{Z}_p[G_n]\to \Bbb{Z}_p[G_n^{p-1}]$$