# Multiplicative Group of $p$ - adic Power Series

We let $$D$$ be an open or closed disc in $$\mathbb{C}_p$$ centered at $$0$$. I have to prove that

$$\text{(a)}$$ $$\{f \in 1 + x\mathbb{C}_p[[x]] : f \text{ converges in }D\}$$ is closed with respect to multiplication but does not form a group; and

$$\text{(b)}$$ $$\{f = 1 + \sum_{i=1}^{\infty} a_i x^i : \text{ord}_p a_i − λi > 0\ \forall i \text{, and tends to }\infty\text{ as }i \rightarrow \infty\}$$ forms a group with respect to multiplication, where $$\lambda \in \mathbb{R}$$ is fixed.

While showing that these sets are closed with respect to multiplication is straightforward, I don't know how to show that $$\text{(a)}$$ is not a group. Clearly it has identity, but how do I show that it's candidate for inverse (which will exist since the constant term is $$1$$) doesn't converge in the disk?

Coming to $$\text{(b)}$$, I'm unable to show how a candidate for inverse's coefficients will satisfy the given constrain, may be mathematical induction will help. But not getting anywhere.

I know asking multiple questions in a single question is frowned upon, but all these questions come from a single question, which is an exercise problem in Koblitz's book on $$p$$ - adic analysis, but here's another question I have which I can't make heads or tails of:

Let $$f_j ∈ 1+x\mathbb{Z}_p[[x]]$$ converge in the closed unit disc $$D(1)$$ for any $$j$$. Does $$f := \prod_{j \geq 1}f_j (x^j)$$ converge in $$D(1)$$?

• With $B = \{ \|x \| < r \}$ don't we have $\{f \in 1 + x\mathbb{C}_p[[x]] : f \text{ converges and has no zeros in } B\}$ is a group. If $f$ has a zero in $B$ its inverse doesn't converge in $B$. And $ord_pa_i \ge \lambda i$ in $(b)$ would be the condition for $f$ having no zeros while $ord_pa_i > \lambda i$ would be the extension to the closed disk. – reuns Nov 9 '18 at 0:55