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This question is an extension of this one.

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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Leo
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    Abelianizing and tensoring with $\mathbb{Q}$ still works. – Qiaochu Yuan Aug 09 '11 at 14:38
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    @Qiaochu: Hmm, I'm confused. Abelianizing what? $F_Y$? How do I prove $F_Y=F_X$? – Leo Aug 09 '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 09 '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 $|Y|$? – Leo Aug 09 '11 at 14:56

1 Answers1

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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.

Qiaochu Yuan
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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? – Leo Aug 09 '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 09 '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. – Leo Aug 09 '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 09 '11 at 15:52
  • thanks man, I appreciate it. And sorry for my confusions... Oh, and just one more thing, do we really have to go to vector spaces (i.e. tensoring with $\mathbb{Q}$), or do free modules provide the answer? Can there be a surjection from a free module with smaller rank to a free module with greater rank? – Leo Aug 09 '11 at 15:58
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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 09 '11 at 16:08