Are the real numbers a subgroup of the circle group? Both the real numbers $\mathbb R$ and the circle group $\mathbb T$ have cardinality continuum.
It is easy to show that $\mathbb T$ is not a subgroup of the additive group of real numbers $\mathbb R(+)$ since $\mathbb T$ has an element $x \ne 0$ such that $x + x = 0$.
Wikipedia states that $\mathbb T \cong \mathbb R \oplus (\mathbb Q/\mathbb Z)$.
(https://en.wikipedia.org/wiki/Circle_group#Group_structure)
Does it mean $\mathbb R(+)$ is isomorphic to a subgroup of $\mathbb T$?
What would be an expression for the isomorphism?
Is it order-preserving for the linear order of $\mathbb R$ and/or for the cyclic order of $\mathbb T$?
(https://en.wikipedia.org/wiki/Cyclic_order)
 A: Yes, it does mean that $\mathbb{R}$ is (isomorphic to) a subgroup of $\mathbb{T}$. You should understand the isomorphism $\mathbb{T}\simeq \mathbb{R}\oplus \mathbb{Q}/\mathbb{Z}$ if you want to see how exactly the identification is made.
But here is the idea of the wikipedia article. Divisible abelian groups are the same as injective $\mathbb{Z}$-modules. An injective abelian group $A$ has the property that, if $A\leq A'$, then $A'=A\oplus B$ for some $B$ (every inclusion splits). If $A$ is a divisible abelian group, so is its torsion subgroup $T:=\mathrm{Tor}(A)$, and therefore the inclusion $T\leq A$ splits: we can write $$A=T\oplus B$$ for some $B$. Now $B\simeq A/T$ is torsion-free, which means it can be viewed as a $\mathbb{Q}$-vector space. Pick a basis for it, so that $B\simeq \mathbb{Q}^I$.
Here's the magic part. If you set $A=\mathbb{T}$, then $A$ is uncountable; but the torsion subgroup $T$ is just the set of all roots of unity, which is countable. Since $A=T\oplus B$, it follows that $B$ must be uncountable. But then the index set $I$ must be uncountable, so $B\simeq \mathbb{Q}^I\simeq \mathbb{R}$ (notice $\mathbb{R}$ has uncountable dimension as a $\mathbb{Q}$-vector space).
Since the group $T$ of roots of unity is isomorphic to $\mathbb{Q}/\mathbb{Z}$, in total we get
$$\mathbb{T}\simeq \mathbb{Q}/\mathbb{Z}\oplus \mathbb{R}.$$
The submodule $\mathbb{R}$ is realized non-canonically: you have to pick a basis for $A/T$.
