(Assuming $\mathfrak{p}>\omega_1$) If G is a separable metrizable group with at least two elements, then $G^{\omega_1}$ is a separable Fréchet group which is not metrizable.
This is stated without proof in the survey article by Justin Tatch Moore and Stevo Todorcevic in Open Problems in Topology 2 titled The metrization problem for Fréchet groups.
I've never studied topological groups before and I have some questions.
How am I supposed to interpret $G^{\omega_1}$? Are its elements finite support $\omega_1$-sequences (almost always $1_G$) or just all $\omega_1$-sequences?
It would also be very useful to me if you could help me verify the claim. It has been suggested to me that this claim might actually be false and a mistake by the authors. (But this might be because of a wrong interpretation of $G^{\omega_1}$.) I'm wondering which of the three properties (separable, fréchet, non-metrizable) are always true and which need the assumption of $\mathfrak{p}>\omega_1$.
Thank you.
EDIT: I've found the following claim which is probably what Moore and Todorcevic intended to convey with their example:
Assuming $\mathfrak{p}>\omega_1$, every countable dense subgroup of $2^{\omega_1}$ is Fréchet and not metrizable.
This subgroup would be Fréchet and separable (both simply because it's countable), but how can I verify it's not metrizable?