# Isomorphism between $\mathbb{T}$ and $\mathbb{R} \oplus \mathbb{Q}/\mathbb{Z}$.

I'm trying to prove that the circle group $\mathbb{T}$ is isomorphic to $\mathbb{R} \oplus \mathbb{Q}/\mathbb{Z}$ with a little bit of cardinal arithmetics. First, I know that $\mathbb{T}$ can be decomposed (structure theorem of divisible groups) as a direct sum of $\mathbb{Q}^X$ for some set $X$ and its torsion subgroup (which is isomorphic to $\mathbb{Q}/\mathbb{Z}$).

I'd like to say that $|X| = |\mathbb{R}|$ (thus $\mathbb{Q}^X \cong \mathbb{R}$ as $\mathbb{Q}$ vector spaces and the desired result follows), however, I don't see how can I show this.

My reasoning was: $|\mathbb{R}|=|\mathbb{T}|=|\mathbb{Q}^X||\mathbb{Q}/\mathbb{Z}|$, then it must be the case that $|\mathbb{T}|=|\mathbb{Q}^X|$ and $|\mathbb{R}|=|\mathbb{Q}^X|$, and here I'm stuck because I cannot conclude that $|X|$ must be $|\mathbb{R}|$ (as counterexample, $|\mathbb{R}| = |\mathbb{Q}^{\mathbb{N}}|$).

Is there a way to prove that $|X| = |\mathbb{R}|$ or should I try another reasoning not involving cardinals?

Edit: forgot $\mathbb{Q}^{X}$ is a direct sum of copies of $\mathbb{Q}$ (is not the whole direct product because $X$ cannot be finite), so my "counterexample" is useless but still cannot see the equality.

• To denote a direct sum of $X$ copies of $Q$, you write $Q^{(X)}$. – user49640 Jun 11 '17 at 2:19
• I haven't looked at your approach in detail, but I would suggest this alternative. We can consider $T$ to be $R/Z$. Let $(e_i)_{i \in I}$ be a basis for $R$ over $Q$, with $e_{i_0} = 1$. Now $R/Z = Qe_{i_0}/Ze_{i_0} \oplus \bigoplus_{i \in I - \{i_0\}} Qe_i$. Since $I$ and $I - \{i_0\}$ have the same cardinality, the factor on the right is isomorphic to $R$. – user49640 Jun 11 '17 at 2:25

I'm not sure how to make it work via cardinal arithmetic, but here is a direct proof.

First, a vector spaces over $\mathbb{Q}$, both $\mathbb{R}$ and $\mathbb{R}\oplus\mathbb{Q}$ have dimension $|\mathbb{R}|$, so they are isomorphic.

Choose an isomoprhism $f:\mathbb{R}\rightarrow \mathbb{R}\oplus\mathbb{Q}$ once and for all. Let $z= f^{-1}(0,1)$. Since $f$ is a homormorphism, $z\neq 0$, so $\langle z\rangle$ is isomorphic to $\mathbb{Z}$.

Then $f$ induces an isomorphism between $\mathbb{R}/\langle z\rangle \cong \mathbb{R}/\mathbb{Z}\cong S^1$ and $\mathbb{R}\oplus \mathbb{Q}/\langle (0,1)\rangle = \mathbb{R}\oplus (\mathbb{Q}/\mathbb{Z})$.

This simple result is needed.

Lemma. Let $V$ be an infinite dimensional vector space defined over the field $K$ with $|K|=\aleph_0$, then $dim_K V = |V|$.

Because $|\mathbb{Q}| = \aleph_0$, $X$ must be an infinite set, $|\mathbb{R}| = |\mathbb{Q}^{(X)}|$, and $\mathbb{R}$ is an infinite dimensional $\mathbb{Q}$-vector space, $\mathbb{R} \cong \mathbb{Q}^{(X)}$ as vector spaces, therefore, $|\mathbb{R}| = |X|$.

The torsion part of a divisible group is itself divisible, so it is a direct summand. The torsion part of $\mathbb{R}/\mathbb{Z}$ is $\mathbb{Q}/\mathbb{Z}$.

Consider a complement $H$, so $$\mathbb{R}/\mathbb{Z}=H\oplus\mathbb{Q}/\mathbb{Z}$$ Then $H$ is a torsion free divisible group, hence a $\mathbb{Q}$-vector space. The dimension of $H$, by cardinality reasons, is $2^{\aleph_0}$, the same as the dimension of $\mathbb{R}$ over $\mathbb{Q}$.