# J. J. Rotman's proof that two free groups are isomorphic iff they have the same rank

Rotman's "An Introduction to the Theory of Groups" contains the above result as Theorem 11.3. However, I failed to pickup a step of the proof. It goes something like this:

($$F \simeq G \implies \operatorname{rank}(F) = \operatorname{rank}(G)$$)

Let $$X$$ be a basis of $$F$$ and let $$Y$$ be a basis of $$G$$. Since $$F \simeq G$$, then $$F/F' \simeq G/G'$$, where $$F'$$ denotes the commutator subgroup. By a previous result, $$F/F'$$ is a free abelian group of basis $$\overline{X} = \{xF'\mid x \in X\}$$, and, by a previous result on free abelian groups, $$|\overline{X}| = |\overline{Y}|$$. As $$\mathbf{|X| = |\overline{X}|}$$, we have the result.

The part in bold is where I couldn't understand. In principle, based on Rotman's definition of a free group with basis $$X$$, I couldn't see a reason why, if $$xF' = \tilde{x}F'$$, then $$x = \tilde{x}$$. In fact, even if I could, at this point, use that $$F$$ is generated by $$X$$, I don't think I'd be able to prove this tiny part of the result...

Could anyone give me any hints as per how to procede?

PS: Rotman's definition of a free group

Def: A group $$F$$ is called free with basis $$\mathbf{X}$$ $$\iff$$ for every group $$G$$ and for every function $$f: X \to G$$, there exists one, and only one, homomorphism $$\phi: F \to G$$ that extends $$f$$.

• Sorry, ignore my previous comment (now deleted). The bijection $X \to \bar{X}$ is indeed $x \mapsto xF'$. This follows because $xy^{-1} \not\in F'$ for $x \ne y \in X$. – Derek Holt May 17 at 17:24
• @DerekHolt No problem! In that case, I'll delete my own, and ask you: how did you prove the assertion in your current comment? – Gauss May 17 at 17:27

As Derek Holt points out, this boils down to showing that $$xy^{-1}$$ is not in the commutator subgroup $$F' \leq F$$ when $$x\neq y$$.
The easiest way to see this is to pick any abelian group $$C$$ with two distinct elements $$a,b$$ (the cyclic group of order $$2$$ will do). Because $$F$$ is free on $$X$$, there exists a homomorphism $$f:F\to C$$ with $$f(x)=a$$ and $$f(y)=b$$. In particular, $$xy^{-1} \notin \ker{f}$$.
But, since $$C$$ is abelian, $$F'\subset \ker{f}$$.