Finding the number of elements of a quotient group.

Question

For a finite extension $K$ of $\mathbb{Q}_p$ with residue field of order $q$, ring of integers $\mathcal{O}_K$, and prime element $\pi$, define $(1+\pi^n\mathcal{O}_K)$ to be the multiplicative subgroup of $\mathcal{O}_K^\times$ constituting the elements $a \in \mathcal{O}_K$ with $a \equiv 1$ (mod $\pi^n)$. Then, how can I show that $$ \#\left (\frac{(1+\pi\mathcal{O}_K)}{(1+\pi^n\mathcal{O}_K)} \right) = q^{n-1} $$ and that $$ \#\left(\frac{\mathcal{O}_K}{\pi^n\mathcal{O}_K}\right)^\times = (q-1)q^{n-1}\,? $$ Presumably, both are shown using similar arguments, but I don't seem to be able to figure out what the line of reasoning is.

Answer 1

Easiest to do it in short steps.

You probably know that $\mathscr O^\times/(1+\pi\mathscr O)$ is a group of order $q-1$, and that this group may also be seen as $\left(\mathscr O/\pi\mathscr O\right)^\times$. That’s the first step, which you show by establishing an isomorphism between this group and $\kappa^\times$, where $\kappa$ is the residue field.

Second step is what you use in an induction, namely that the multiplicative group $(1+\pi^m\mathscr O)/(1+\pi^{m+1}\mathscr O)$ is isomorphic to the additive group $\kappa^+$. Do you see the isomorphism? It is to take $1+\pi^mz$ and send it to $z\pmod{\pi\mathscr O}$. Of course you have to verify that the map really is a homomorphism, with kernel $1+\pi^{m+1}\mathscr O$. I’ll leave the details to you, but get back in a comment if you have trouble.