free groups: $F_X\cong F_Y\Rightarrow|X|=|Y|$ I'm reading Grillet's Abstract Algebra. Let $F_X$ denote the free group on the set $X$. I noticed on wiki the claim $$F_X\cong\!\!F_Y\Leftrightarrow|X|=|Y|.$$ How can I prove the right implication (find a bijection $f:X\rightarrow Y$), i.e. that the rank is an invariant of free groups?
I am hoping for a simple and short proof, having all the tools of Grillet at hand. Rotman (Advanced Modern Algebra, p.305) proves it only for $|X|<\infty$, Bogopolski's (Introduction to Group Theory, p.55) proof seems (unnecessarily?) complicated, and Lyndon & Schupp's (Combinatorial Group Theory, p.1) proof I don't yet understand. It's the very first proposition in the book; in the proof, they say:

The subgroup $N$ of $F$ generated by
  all squares of elements in $F$ is
  normal, and $F/N$ is an elementary
  abelian $2$-group of rank $|X|$. (If
  $X$ is finite, $|F/N|=2^{|X|}$ finite;
  if $|X|$ is infinite, $|F/N|=|X|$). $\square$

Is $N:=\langle w^2;w\in F\rangle$? What is an abelian $2$-group? Elementary? What and how does the above quote really prove? I'm guessing a free abelian group on $X$ is $\langle X|[X,X]\rangle\cong\bigoplus\limits_{x\in X} \mathbb{Z}$?
Can an isomorphism $\varphi:F_X\rightarrow F_Y$ not preserve the length of words? At least one letter words?

The first part of your question is relatively easy to get. say f is the homomorphism between the two free groups then you need to deduce that f induce a bijection (as a function not a homomorphism) between elements of X and a free (no obvious combinations) set of Fy (which of the same cardinality as Y) let us suppose that f(X) is not a free set in Fy i.e there are a1,a2,..ak in (f(X) and f(X)^(-1)) such that a1a2..ak=1  by f-1 you will end up having a relation in Fx which is supposed to be free so you have |Y|<|X| the rest follows..
