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?

  • 5
    $\begingroup$ 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. $\endgroup$ – Derek Holt Feb 17 '19 at 16:39

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 $i=1,\dots, n$.


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