Proving the fact "cardinality of a basis of free Abelian group is unique"

Question

I have a question about the proof of theorem 7.3 in Serge Lang's book.

It says, cardinality of a basis of the free Abelian group is unique.

For that, suppose B and C are cardinalities of such bases of G. Then,

$2^B = |G/2G| = 2^C$

so B = C.

However, according to the most popular answer of https://mathoverflow.net/questions/1924/what-are-some-reasonable-sounding-statements-that-are-independent-of-zfc?page=1&tab=votes#tab-top , it is not valid in ZFC to assert

$B < C \rightarrow 2^B < 2^C$

Then how can we prove the following?

$2^B = 2^C \rightarrow B = C$

Thank you.

Answer 1

$G/2G$ is more than just an abelian group with $2^B$ ($=2^C$) elements, it is also an $\Bbb F_2$ vector space with $B$ (or $C$) as a basis. So in order to conclude that $|B|=|C|$, we do not (illegally) use $|2^B|=|2^C|\to |B|=|C|$, but rather that the cardinality of a vector space basis (aka. dimension) is unique.