Every Lindelöf topological group is isomorphic to a subgroup of the product of second countable topological groups. I want to show that every Lindelöf topological group is isomorphic to a subgroup of the product of second countable topological groups. I received an answer using the fact that Lindelöf topological groups are $\omega$-narrow, but I want to show it by using the following theorem.
Theorem: Every Hausdorff topological group $G$ is topologically isomorphic to a subgroup of the group of isometries $Is(M)$ of some metric space $M$, where $Is(M)$ is taken with the topology of pointwise convergence.
Any help would be greatly appreciated!
 A: A nice reference for this sort of thing is the book Topological Groups and Related Structures by Arhangel'skii and Tkachenko.
A Hausdorff topological group $G$ is said to be $\omega$-narrow if for every open neighborhood $U$ of the identity $e$, there is a countable set $A$ such that $AU=G$.
Certainly every Lindelöf topological group is $\omega$-narrow; take $A\subset G$ to be a countable set such that $\{aU\}_{a\in A}$ is an open cover of $G$.
Guran's Theorem (3.4.23 in the book mentioned) states that a topological group is $\omega$-narrow iff it embeds as a topological subgroup of a product of second countable topological groups. 
This result is more general than the one you are asking for and the proof can be found in the book. On the other hand, the proof here doesn't seem to use Uspenskij's theorem (that $G$ can be embedded in the isometry group of some metric space $M$, in particular the metric space of all bounded left uniformly continuous real-valued functions on $G$).
Perhaps for Lindelöf $G$, there is a simpler proof using Uspenskij's theorem and someone else can point the way to this. I am curious to know where it is said that such a proof is possible?
A: Let $G$ be a Lindelöf group. By your theorem, you can assume that $G$ is a subgroup of $\operatorname{Iso}(M)$ for some metric space $M$. Now, consider the decomposition of $M$ into $G$-orbits, call them $M_a$. Since each orbit is an image of $G$, each $M_a$ is Lindelöf. Furthermore, since each $M_a$ is metrizable, they each have a countable base. 
Now, for every $M_a$, since $G$ acts on $M_a$, there is a natural homomorphism of topological groups from $G$ to $\operatorname{Iso}(M)$. Then, the diagonal product of these homomorphisms gives you the embedding of $G$ into the product of second countable groups.
