I am trying to understand the following (from Munkres)
Thm: If G, a free abelian group, has basis $\{a_1,...,a_n\}$, then $n$ is uniquely determined by G.
What does this mean ? What exactly is being determined ? Does it mean we can find n if we know G ?
The proof finishes by saying that the cardinality of $G/2G$ is $2^n$. I don't exactly understand how it helps us.
Thanks.
Yes, it's saying that if you know $G$ then you can find $n$.
It's possible that a module might have bases of different sizes. However, this theorem tells you that all bases of a given free abelian group have the same size. Another way of wording it would be to say that if $a_1, \dots, a_n$ and $b_1, \dots, b_m$ are bases of $G$ then $m=n$.
This is much like vector spaces, all of whose bases have the same size (which is precisely the dimension of the space): if $e_1, \dots, e_n$ and $e'_1, \dots, e'_m$ are bases of the same vector space $V$ then $m=n=\dim V$.
Regarding the last part of your question, if you know that $|G/2G|=2^n$ then $$n = \log_2 |G/2G|$$ Notice how this formula doesn't mention any of the $a_i$ $-$ it's independent of the basis! So you can get $n$ just by knowing $G$, without knowing the specific basis you're working with.
Edit (following your comment):
The simplest nontrivial example is $\mathbb{Z}$ (under addition); the only two possible bases are $\{ 1 \}$ and $\{ -1 \}$, each having size $1$.
More generally, $\mathbb{Z}^n$ is a free abelian group for each natural number $n$, and each basis of $\mathbb{Z}^n$ has size $n$. For example, possible bases of $\mathbb{Z}^2$ include: $$\{ (1,0), (0,1) \}, \{ (1,0), (0,-1) \}, \{ (1,0), (1,1) \}$$ or even $\{ (1,0), (k,1) \}$ for any $k \in \mathbb{Z}$.
And indeed $$\mathbb{Z}^n / 2 \mathbb{Z}^n = \{ (i_1, i_2, \dots, i_n)\, :\, i_k=0\ \text{or}\ 1\ \text{for each}\ k \}$$ which has size $2^n$.