3
$\begingroup$

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?

$\endgroup$
1
  • 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
2
$\begingroup$

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 $i=1,\dots, n$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.