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 , , and are all isomorphic.

Indeed, they are all -vector spaces of dimension . Another less known example is:

**Theorem:** and are isomorphic.

**Proof.** Let be a basis of as a -vector space; without loss of generality, suppose . We can write as a disjoint union where and . Let and denote the subspaces generated by and respectively; as additive groups, they are both isomorphic to since they are -vector spaces of dimension . Finally, let denote the subspace generated by . Then

.

Now, notice that the morphism

gives an isomorphism , and the morphism

gives an isomorphism . Therefore, we deduce that

.

Noticing that the polar decomposition

gives an isomorphism , we conclude the proof.

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: and give such an example as -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 , 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 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.

### Like this:

Like Loading...

*Related*