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?
Best Answer
It only says at the hint of the exercise, but probably you're required to prove the universal property with respect to Abelian groups:
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 $i=1,\dots, n$.