# Proving the Free Abelian Group is Free Abelian…?

On page 40 of these notes is the following exercise:

Prove that the group with generators $$a_1,...,a_n$$ and relations $$[a_i,a_j]=1$$, $$i \neq j$$, is the free abelian group on $$a_1,...,a_n$$.

On page 35 is the following definition:

The free abelian group on generators $$a_1,...,a_n$$ has generators $$a_1,...,a_n$$ and relations $$[a_i,a_j]$$, $$i \neq j$$.

I'm a little puzzled. What exactly is there to prove?

• I agree. The author has asked you to prove something that is true by definition. There are a number of different and equivalent definitions of free abelian groups so I expect they forgot which one they had used. – Derek Holt Feb 17 '19 at 16:39

Let $$F:=\langle a_1,\dots, a_n\mid [a_i, a_j] =1\rangle$$, and let $$A$$ be an arbitrary Abelian group, with an evaluation map $$f:\{a_1,\dots, a_n\} \to A$$.
You have to prove that there is a unique homomorphism (of Abelian groups) $$\tilde f:F\to A$$ such that $$\tilde f(a_i)=f(a_i)$$ for each $$i=1,\dots, n$$.