2
$\begingroup$

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 $|X|$?

  • 0
    @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 $|Y|$?2011-08-09

1 Answers 1

6

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.

  • 2
    @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](http://en.wikipedia.org/wiki/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.2011-08-09