# 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)

• It looks like they're arguing that each element $z$ of $\mathbb{T}$ (other than the roots of unity) together with its divisors (elements so that $nw = z$ for some integer $n$) generate a subgroup isomorphic to $\mathbb{Q}$. By cardinality reasons, the number of such subgroups should be countable. But up to isomorphism there's a unique vector space of a certain cardinality over a fixed field; it happens that $\mathbb{R}$ is a vector space over $\mathbb{Q}$ of cardinality of the continuum. So abstractly there's an isomorphism of the torsion-free part of $\mathbb{T}$ with $\mathbb{R}$ (con't) – Jane Doé May 4 at 5:04
• as $\mathbb{Q}$-vector spaces. However, an explicit isomorphism of $\mathbb{R}$ with $\oplus_\mathfrak{c} \mathbb{Q}$ is difficult to write down... it's known as the problem of finding a Hamel basis for $\mathbb{R}$ over $\mathbb{Q}$. So yes, $\mathbb{R},+$ is isomorphic to a subgroup of $\mathbb{T}$, but the isomorphism is not naively of geometric origin I would say. – Jane Doé May 4 at 5:05

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 $$A/T$$.
• Is it an order-preserving isomorphism for the cyclic order of $\mathbb T$ and/or for the linear order of $\mathbb R$? – Alex C May 4 at 9:48
• I've found in G. G. Pestov work: $\mathbb{T} \simeq C \oplus B$, where $C$ is the group of all roots of unity, and $B$ is a linearly ordered group of cardinality continuum. $B$ is dense, but not continuous, so it is not isomorphic to $\mathbb R$ considering continuity. Is this correct? – Alex C May 4 at 15:38