I am trying to know whether the following result is true.

Let $F$ be a free group with a basis $X$, and let $X'=\{xF':x\in X\}$, where $F'$ is the commutator subgroup of $F$. Then, $|X|=|X'|$.

In order to answer this question, I am trying to prove that $x\mapsto xF'$ is a bijection. It is clearly a surjection. So, I am trying to prove that $x_1F'=x_2F'$ implies that $x_1=x_2$. But, I really can't move on.

I hope that this question is not too stupid. Any help?

  • $\begingroup$ When you say that $F$ is generated by $X$, do you mean that $X$ is a free basis for $F$? $\endgroup$ – James Jul 23 '14 at 18:41
  • $\begingroup$ @James Yes, $X$ is a basis of $F$. $\endgroup$ – YYF Jul 23 '14 at 18:42

Note that $xF' = yF'$ if and only if $y^{-1}x\in F'$. By the universal property of the commutator subgroup, any morphism $\varphi$ of $F$ into an abelian group factors uniquely through $F/F'$.

That is, given $\varphi: F \rightarrow A$ for $A$ abelian, there exists a unique $\bar\varphi: F/F' \rightarrow A$ so that $\bar\varphi \circ \pi =\varphi$, where $\pi$ is the usual quotient homomorphism:

$$\require{AMScd} \begin{CD} F @>\varphi>> A \\ @VV\pi V @V \mathbb 1 VV \\ {F/F'} @>\bar\varphi>> A \end{CD}$$

If $y^{-1}x \in F'$, it follows that any homomorphism $\varphi$ of $F$ into an abelian group takes $y^{-1}x$ to the identity.

For any distinct free generators $x$ and $y$ of the free group, can you construct a homomorphism of $F$ into an abelian group that takes $y^{-1}x$ to a nonzero element?

  • $\begingroup$ Is there another way to not use the universal property of the commutator subgroups? I have not learned it yet and am really wondering why the author did not introduce such a useful property and just waved his hand to claim that $|X|=|X'|$. $\endgroup$ – YYF Jul 23 '14 at 18:47
  • $\begingroup$ I'm sure you could give an argument working with directly with elements. But proving the needed property is easy -- maybe stating it as a universal property made it sound more complex than it is. You don't need uniqueness, just existence of $\bar\varphi$ (with notation as above). If $\varphi: F \rightarrow A$ is a homomorphism of $F$ into an abelian group, elements of the form $xyx^{-1}y^{-1}$ -- generators of $F'$-- are sent to the identity. Hence $F' \subset \ker \varphi$, and the rule $xF' \mapsto \varphi(x)$ gives a well-defined homomorphism from $F/F'$ to $A$ with the desired property. $\endgroup$ – vociferous_rutabaga Jul 23 '14 at 18:58
  • $\begingroup$ You don't need the full universality property. If you can map distinct basis elements to distinct elements in some abelian group, then the kernel of the homomorphism contains $F^{\prime}$. And, since those cosets of the kernel are distinct, so are the cosets of $F^{\prime}$. $\endgroup$ – James Jul 23 '14 at 18:59
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    $\begingroup$ +1 for emphasizing the usefulness of universal properties! $\endgroup$ – Jakob Werner Jul 23 '14 at 20:06

I like the answer from Morgan O, but you could also get this from the following fact, which you may already know (and is not too hard to prove): the derived subgroup $F^{\prime}$ consists of those elements of $F$, thought of as (reduced) words on $X$, for which the exponent sum of each generator from $X$ is equal to zero.

Now, if $x,y\in X$ and $xF^{\prime} = yF^{\prime}$, then $x^{-1}y\in F^{\prime}$, so the exponent sum of $x$ in the word $x^{-1}y$ can only be zero if $y=x$.


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