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 $\mathbb T \cong \mathbb R \oplus (\mathbb Q/\mathbb Z)$.

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 $\mathbb T$?

  • $\begingroup$ 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) $\endgroup$ – Jane Doé May 4 at 5:04
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    $\begingroup$ 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. $\endgroup$ – 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 $A/T$.

  • $\begingroup$ Is it an order-preserving isomorphism for the cyclic order of $\mathbb T$ and/or for the linear order of $\mathbb R$? $\endgroup$ – Alex C May 4 at 9:48
  • $\begingroup$ The construction is highly non-canonical, so I suggest you examine it carefully to see if it preserves additional properties. Unfortunately I don't know what "cyclic order" means. $\endgroup$ – Ehsaan May 4 at 15:17
  • $\begingroup$ 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? $\endgroup$ – Alex C May 4 at 15:38
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    $\begingroup$ That is correct, although your question was not about continuity. $\endgroup$ – Lee Mosher May 10 at 19:34

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