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?

Yes, $N$ is the subgroup generated by all squares. $N$ is normal in $F$, because if $w^2$ is an element of the generating set for $N$, and $x\in F$, then $xw^2x^{-1} = (xwx^{-1})^2$ is also an element of $N$. So for every $x\in F$ we have $xNx^{-1}\subseteq N$, hence $N$ is normal.

When $p$ is a prime, an abelian $p$-group is simply an abelian group all of whose elements have order a power of $p$ (the group could be finite or infinite). An elementary abelian $p$-group is an abelian $p$-group in which every element satisfies $a^p = 1$ (and so, every element except for the identity is of order exactly $p$; this is with multiplicative notation, you would have $pa=0$ if you are using additive notation for your group). So the assertion is that $F/N$ is abelian, and the square of every element is the identity.

The fact that the square of every element of $F/N$ is the identity follows because every square is in $N$: if $fN\in F/N$, then $(fN)^2 = f^2N = 1N$. And the fact that $F/N$ is abelian now follows from the well-trod fact that a group in which the square of every element is the identity must be abelian (since $1 = (ab)^2 = a^2b^2$, so $abab=aabb$, hence $ba=ab$ by cancellation).

An elementary abelian $p$-group is always a vector space over $\mathbb{F}_p$, the field with $p$-elements: given $\alpha\in\mathbb{F}_p$, let $a\in\mathbb{Z}$ by any integer mapping to $\alpha$. Then, assuming your group is written additively, define $\alpha\cdot g$ as $\alpha\cdot g= ag$. Since $pg=0$, this is well defined and makes the abelian group into a vector space.

So here, you have that $F/N$ is an abelian $2$-group, and therefore is a vector space over the field of $2$-elements.

In fact, the images of the free generating set $X$ in $F/N$ form a basis for this vector space: since $X$ spans $F$, its images span $F/N$. And if you have a nontrivial linear combination between them, then it must be of the form $$\overline{x_1}+\cdots + \overline{x_n} = \mathbf{0}$$ where $\overline{g}$ is the image of $g\in F$ in the quotient, and $x_1,\ldots,x_n$ are pairwise distinct elements of $X$. But this means that $x_1\cdots x_n\in N$, that is, that it is the square of an element of $F$, and this is easily shown to be impossible.

So $F/N$ is a vector space over $\mathbb{F}_2$, and has a basis of cardinality $|X|$. How many elements does a vector space of dimension $\kappa$ over $\mathbb{F}_2$ have? If $\kappa$ is finite, then it has $2^{\kappa}$ elements. If $\kappa$ is infinite, then it has $\kappa$ elements. So if $X$ is infinite, then $|X|=|F/N|$.

Now suppose that $F_X$ and $F_Y$ are isomorphic. Then the isomorphism maps $N=\langle w^2\mid w\in F_X\rangle$ to $M=\langle z^2\mid z\in F_Y\rangle$, so we get that $F_X/N\cong F_Y/M$. If one of them is finite, then they both are; if one of them is infinite, then they both are. If both are finite, then the cardinality of $F_X/N$ is $2^{|X|}$, and the cardinality of $F_Y/M$ is $2^{|Y|}$, and since they are isomorphic groups, then $|X|=|Y|$. If they are both infinite, then $F_X/N$ has cardinality $|X|$, and $F_Y/M$ has cardinality $|Y|$, and since they are isomorphic their cardinalities are the same, so $|X|=|Y|$ as well. This proves the result.

There is a simpler way: the result holds for abelian free groups (tensor up to $\mathbb{Q}$ to reduce to the vector space $K$), and then show that $F_X^{\mathrm{ab}}$ is the free abelian group on $X$.

To answer final question: "word length" depends on the free basis. There is always a choice of basis for $F_Y$ that makes $\varphi$ preserve word length (simply take the basis $\varphi(X)$), but in general it need not. Take $X=\{x_1,x_2\}$, $Y=\{y_1,y_2\}$, and first map $F_X$ to $F_Y$ the obvious way ($x_1\mapsto y_1$, $x_2\mapsto y_2$), and then compose with a suitable inner automorphism of $F_Y$. For example, composing with conjugation by $y_1$ maps $x_1\mapsto y_1$ and $x_2\mapsto y_1y_2y_1^{-1}$. Composing with conjugation by $y_1y_2y_1^{-1}$ makes it even worse: $$\begin{align*} x_1 &\mapsto (y_1y_2y_1^{-1})y_1(y_1y_2^{-1}y_1^{-1}) = y_1y_2y_1y_2^{-1}y_1^{-1},\\ x_2 &\mapsto (y_1y_2y_1^{-1})y_2(y_1y_2^{-1}y_1^{-1}) = y_1y_2y_1^{-1}y_2y_1y_2^{-1}y_1^{-1}, \end{align*}$$ mapping the two generators to words of length 5 and 7, respectively; of course, you can make it pretty much as bad as you want using this idea.

Let $G$ be some group of order $n$ where $1\lt n\lt\aleph_0$. If $F_X\cong F_Y$, then the number of homomorphisms from $F_X$ to $G$ is equal to the number of homomorphisms from $F_Y$ to $G$, that is, $n^{|X|}=n^{|Y|}$. If $\min(|X|,|Y|)$ is finite, it follows from $n^{|X|}=n^{|Y|}$ that $|X|=|Y|$. If $|X|$ and $|Y|$ are both infinite, then $|X|=|F_X|=|F_Y|=|Y|$.