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.