I am reading a proof with the followings setup and claim.
$K/F$ is a Galois extension of local fields with group $G$ of order $n = q^s$, where $q$ is prime and $s \geq 1$. Assume the maximal ideal $\mathfrak{p}_F$ in $\mathcal{O}_F$ totally ramifies in $K$ (so that $\mathfrak{p}_F\mathcal{O}_K = \mathfrak{p}^n$ where $\mathfrak{p}$ is the maximal ideal in $\mathcal{O}_K$), and $\mathfrak{p}_F \nmid n$.
If $U = \mathcal{O}_K^\times$ and $U^{(n)} = 1 + \mathfrak{p}^n$, then the author states that "$U^{(1)}$ is the inverse limit of finite $p$-groups, $\mathfrak{p}_F | p$, so is uniquely divisible by $n$", which is the statement that I can't understand. I see that $$U^{(1)} = \varprojlim_{n > 1} U^{(1)}/U^{(n)}.$$ Since $U/U^{(n)} \cong (\mathcal{O}_K/\mathfrak{p}^n)^\times$, I guess it would suffice to show that these are $p$-groups for $n > 1$, because $U^{(1)}/U^{(n)} \leq U/U^{(n)}$. But I'm not entirely sure why that's true.
But even then, I'm not sure why it follows that $U_1$ is uniquely divisible by $n$, which is prime to $p$. Isn't it true that in general, a finite $p$-group isn't uniquely divisible, let alone divisible, by $n$?
In any case, the point of this is to show that $H^1(G, U^{(1)}) = 0$, so if there's another elementary way of seeing this, that would help as well.
What am I missing here?
Finite $p$-groups are uniquely divisible by any $n$ relatively prime to $p$. The point is that in a finite $p$-group $G$ the map
$$\mathbb{Z} \ni n \mapsto (g \mapsto g^n)$$
extends uniquely to a map from the $p$-adic integers $\widehat{\mathbb{Z}}_p$, and in the $p$-adic integers any $n$ relatively prime to $p$ has a unique multiplicative inverse. (I could use a finite quotient of $\mathbb{Z}$ depending on the size of $G$, but this way of saying things is independent of the size of $G$.)
For example, in a finite $2$-group there are unique cube roots given by raising an element to the power of
$$\frac{1}{3} = 1 - 2 + 2^2 - 2^3 \pm \dots$$