Although algebraic structures (such that groups, rings, fields, etc.) are generally difficult to classify, surprisingly linear algebra tells us that vector spaces (over a fixed field) are classified up to isomorphism by only one number: the dimension! But a vector field gives rise to a group and a vector space isomorphism gives rise to a group isomorphism. Therefore, it is not surprising that linear algebra can sometimes be used in order to verify that some groups are isomorphic. For instance,

Theorem: The additive groups $\mathbb{R}^n$, $\mathbb{C}^n$, $\mathbb{R}[X]$ and $\mathbb{C}[X]$ are all isomorphic.

Indeed, they are all $\mathbb{Q}$-vector spaces of dimension $2^{\aleph_0}$. Another less known example is:

Theorem: $T= \{ z \in \mathbb{C} \mid |z|=1 \}$ and $\mathbb{C}^{\times}$ are isomorphic.

Proof. Let $B$ be a basis of $\mathbb{R}$ as a $\mathbb{Q}$-vector space; without loss of generality, suppose $1 \in B$. We can write $B$ as a disjoint union $B_1 \coprod B_2$ where $|B_1|=|B_2|=|B|$ and $1 \in B_1$. Let $R_1$ and $R_2$ denote the subspaces generated by $B_1$ and $B_2$ respectively; as additive groups, they are both isomorphic to $\mathbb{R}$ since they are $\mathbb{Q}$-vector spaces of dimension $2^{\aleph_0}$. Finally, let $Z$ denote the subspace generated by $1$. Then

$\mathbb{R}/ \mathbb{Z} \simeq (R_1 \oplus R_2) / Z \simeq R_1/ Z \oplus R_2 \simeq \mathbb{R}/ \mathbb{Z} \oplus \mathbb{R}$.

Now, notice that the morphism

$\theta \in \mathbb{R}/ \mathbb{Z} \mapsto e^{i \theta}$

gives an isomorphism $( \mathbb{R} / \mathbb{Z} , +) \simeq ( T, \times )$, and the morphism

$x \in \mathbb{R} \mapsto e^x$

gives an isomorphism $(\mathbb{R},+) \simeq (\mathbb{R}_+, \times)$. Therefore, we deduce that

$T \simeq \mathbb{R} / \mathbb{Z} \simeq \mathbb{R}/ \mathbb{Z} \oplus \mathbb{R} \simeq T \oplus \mathbb{R}_+$.

Noticing that the polar decomposition

$(\theta,r) \in T \oplus \mathbb{R}_+ \mapsto r \cdot e^{i \theta}$

gives an isomorphism $T \oplus \mathbb{R}_+ \simeq \mathbb{C}^+$, we conclude the proof. $\square$

Nevertheless, the converse of our criterium does not hold, that is there exist two non-isomorphic vector spaces such that the associated groups are isomorphic: $\mathbb{R}$ and $\mathbb{R}^2$ give such an example as $\mathbb{R}$-vector spaces. In particular, it is worth noticing that the choice of the field is crucial.

Nota Bene: In this note, we used the axiom of choice dramatically, assuming that every vector space has a basis. In fact, the axiom of choice turns out to be necessary at least to prove the isomorphism $\mathbb{C} \simeq \mathbb{R}$, as noticed by C.J. Ash in his article A consequence of the axiom of choice, J. Austral. Math. Soc. 19 (series A) 1975 (pp. 306-308). More precisely, the author shows that the isomorphism $\mathbb{C} \simeq \mathbb{R}$ implies the existence of a set of reals not Lebesgue measurable, concluding thanks to a model for ZF due to Solovay in which every subset of reals is Lebesgue measurable.