The Sorgenfrey line $\mathbb S$ is the real line endowed with the topology generated by the base consisting of all half-intervals $[a,b)$ for real numbers $a<b$.

The Sorgenfrey line is first-countable and non-metrizable and hence is not homeomorphic to a topological group.

On the other hand, the Sorgenfrey line $\mathbb S$ is homeomorphic to a subset of a topological group. For example, the free topological group $F(\mathbb S)$ over $\mathbb S$ contains a closed topological copy of $\mathbb S$. But $F(\mathbb S)$ also contains a topological copy of the square $\mathbb S\times\mathbb S$ and hence $F(\mathbb S)$ contains an uncountable discrete subspace. Is this situation typical?

Problem. Let $G$ be a topological group containing a topological copy of the Sorgenfrey line. Does $G$ necessarily contain a uncountable discrete subspace?

Added in Edit. The answer to this problem is affirmative under OCA (the Open Coloring Axiom), which follows from PFA (the Proper Forcing Axiom).

Theorem (OCA). Under OCA, each topological group containing a topological copy of the Sorgenfrey line has uncountable spread.

Under PFA we can prove a bit stronger statement.

Theorem (PFA). Under PFA, each topological group $G$ has uncountable spread if $G$ contains a subset, homeomorphic to an uncountable subspace of the Sorgenfrey line.

We recall that a topological space has a countable spread if it does not contain an uncountable discrete subspace.

The proofs of these two results are not very short, so I will add a link to the proof as soon as the corresponding paper will be ready.

So, now the question is about the answer to the above problem in ZFC.