2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ +1 for emphasizing the usefulness of universal properties! $\endgroup$ – Jakob Werner Jul 23 '14 at 20:06
2
$\begingroup$

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 $y=x$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.