Let $1+t\mathbb{F}_q[[t]]=\mathbb{F}_q[[t]]^\times.$
I saw that a similar fact holds as algbras, which seems stronger, but I am wondering if there is an easier proof method if we only care about the group structure. This is because I am working over a much bigger group where this machinery may be useful. I imagine uncoutability will be of use again, but im not exactly sure how.
Any finitely generated group is countable - just enumerate all words in a generating set and their inverses. On the other hand, $G=1+t\mathbb F_p[[t]]$ has the same cardinality as $\mathbb F_p[[t]]$, which is bijective to $(\mathbb F_p)^{\mathbb N}=\operatorname{Maps}(\mathbb N,\mathbb F_p)$, hence uncountable. In fact, the same argument shows that $G$ is not even countably generated, as such groups are also countable.
