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Let $F_X$ denote the free group on the set $X$. For any group $G$ and subset $S\!\subseteq\!G$, $\langle S\rangle$ denotes the subgroup generated by $S$ and $\mathrm{rank}(G) :=\min\{|S|;\:S\!\subseteq\!G, \langle S\rangle\!=\!G\}$.

PROPOSITION:
a) $F_X\cong F_Y\:\Leftrightarrow\:|X|=|Y|$
b) $\mathrm{rank}(F_X)=|X|$

Thus for every cardinal number $c$, there is (up to isomorphism) exactly one free group of rank $c$.

Proof:

a) $(\Leftarrow)$: If $f\!:X\rightarrow Y$ is the bijection, then $\varphi(x_1\ldots x_k):=f(x_1)\ldots f(x_k)$ is the isomorphism.

$(\Rightarrow)$: $F_X\!\cong\!F_Y$ $\Rightarrow$ $\mathrm{Ab} F_X\!\cong\!\mathrm{Ab} F_Y$ $\Rightarrow$ $\oplus_{x\in X}\mathbb{Z}\!\cong\!\oplus_{y\in Y}\mathbb{Z}$, so $|X|\!=\!|Y|$, since rank is known to be an invariant of free modules.

Alternatively, $\big(\oplus_{x\in X}\mathbb{Z}\big)\otimes_\mathbb{Z}\mathbb{Q}$ $\cong$ $\oplus_{x\in X}\big(\mathbb{Z}\otimes_\mathbb{Z}\mathbb{Q}\big)$ $\cong$ $\oplus_{x\in X}\mathbb{Q}$, so $\oplus_{x\in X}\mathbb{Q}$ $\cong$ $\oplus_{y\in Y}\mathbb{Q}$, even as $\mathbb{Q}$-modules, but isomorphic vector spaces are known to have equipollent bases.

b) Since $\langle X\rangle\!=\!F_X$, $\mathrm{rank}(F_X)\leq|X|$. Suppose we have $Y\!\subseteq\!F_X$, $\langle Y\rangle\!=\!F_X$, $|Y|\!<\!|X|$.

QUESTION: how can I finish the proof of b), i.e. prove that $F_X$ can't be generated by a subset with smaller cardinality than $|X|$?

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Abelianizing and tensoring with $\mathbb{Q}$ still works. –  Qiaochu Yuan Aug 9 '11 at 14:38
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@Qiaochu: Hmm, I'm confused. Abelianizing what? $F_Y$? How do I prove $F_Y=F_X$? –  Leon Lampret Aug 9 '11 at 14:48
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You don't. The hypothesis implies that a vector space of dimension $|Y|$ surjects onto a vector space of dimension $|X|$, which is already enough to conclude. –  Qiaochu Yuan Aug 9 '11 at 14:50
    
@Qiaochu: how exactly do I get a vector space? $(\mathrm{Ab}\langle Y\rangle)\otimes_\mathbb{Z}\mathbb{Q}$? How do I know this has rank $|Y|$? –  Leon Lampret Aug 9 '11 at 14:56
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1 Answer

up vote 5 down vote accepted

Here is the argument written out in full so you can tell me which step you don't understand. Suppose $F_X$ is generated by a subset $Y$ with $|Y| < |X|$. This induces a surjection $F_Y \to F_X$. Abelianization gives a surjection $\text{Ab}(F_Y) \to \text{Ab}(F_X)$. Tensoring with $\mathbb{Q}$ gives a surjection from a vector space of dimension $|Y|$ to a vector space of dimension $|X|$; contradiction.

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Ah, now it makes perfect sense. The problem I had was that I didn't know what you were abelianising, $F_Y$ or $\langle Y\rangle$; I constantly had the feeling that I was implicitly assuming the thing I was proving. Just one more detail, the map $F_Y\rightarrow F_X$ is actually tne inclusion, since $Y\subseteq X$, right? –  Leon Lampret Aug 9 '11 at 15:23
    
@Leon: I'm not assuming $Y \subseteq X$; that wouldn't solve the problem. I'm just assuming $Y \subseteq F_X$ generates it. (Ah, I was confused by your notation $\langle Y \rangle$: I thought this meant $F_Y$, but you meant the subgroup of $F_X$ generated by $Y$. Which confuses me, because this is just $F_X$.) –  Qiaochu Yuan Aug 9 '11 at 15:25
    
Ah, $Y\subseteq X$ was stupid remark, sorry. $\langle Y\rangle$ means subgroup generated by $Y$ and $F_Y$ means free group on the alphabet $Y$. But why would $F_Y$ equal $\langle Y\rangle$? After all, $F_X=\langle F_X\rangle$, but $F_{F_X}\ncong F_X$. Anyway, $F_Y$ is a group of words on words on $X$. So the homomorphism $F_Y\to F_Y$ takes words on words on $X$ to words on $X$. I think I get the picture now. –  Leon Lampret Aug 9 '11 at 15:36
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@Leon: I was confused about what $\langle Y \rangle$ meant. And yes, that's right. There's an interesting way to think about how to write down a group structure on a set $G$ related to this: it's equivalent to providing a map $F_G \to G$ satisfying certain requirements spelled out by the axioms of a monad. –  Qiaochu Yuan Aug 9 '11 at 15:52
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@Leon: over general rings, yes. The property that ranks of free modules behave the way you expect them to is known as the invariant basis number property. Obviously every ring that embeds into a division ring (such as $\mathbb{Z}$) has IBN by tensoring. I think it is known that every commutative ring has IBN. But commutative algebra is difficult and I prefer to reduce to linear algebra whenever possible. –  Qiaochu Yuan Aug 9 '11 at 16:08
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