How can I show the free abelian group of rank $r$ is isomorphic to an $r$-copy of $\mathbb{Z}_\infty$?

Question

Today, this problem was given to me.

Let $F$ be a abelian free group of rank $r$. Show that it is isomorphic to an $r$-copy of $Z_{\infty}$.

I could do some messy job about it but so far I failed to solve it. The time for solving it in time is over, but please help me solve it. Thank you very much.

Answer 1

I believe that you mean to say a free abelian group of rank $r$. The definition of a free abelian group $\mathfrak{F}$ of rank $r$ is a group with a generating set $\mathfrak{S}$ of size $r$ for which the only relation is that $[s,t]=1$ for each $s,t\in \mathfrak{S}$. Note that free abelian groups are not free groups when $r\geq 2$.

Once we have this definition digested, the path to victory is quite clear. Let $\epsilon_i$ denote the generator of the $i^{\rm th}$ $\mathbb{Z}$ in $\mathbb{Z}^{r}$. Number the generators in $\mathfrak{S}$ as $s_1,\ldots,s_r$. Define $\Phi:\mathfrak{F}\rightarrow \mathbb{Z}^r$ by $\Phi:s_i\rightarrow \epsilon_i$. Now, can you prove that $\Phi$ is an isomorphism?

We have the relations that $[s_i,s_j]=1$ for each $i,j$. Given an arbitrary word $w=s_{i_1}^{e_1}s_{i_2}^{e_2}\cdots s_{i_s}^{e_s}$ (with $i_1,\ldots,i_s$ not necessarily distinct) can you rewrite $w$ in the form $s_1^{f_1}\cdots s_r^{f_r}$? Where does $\Phi$ send that in $\mathbb{Z}^r$? Finally, can you prove that $\Phi$ is bijective using the fact there are no other relations on $\mathfrak{F}$?