0
$\begingroup$

Let $F[A]$ be the free group on the generating set $A$. Let $C$ be the commutator subgroup of $F[A]$, then show that $F[A]/C$ is a free abelian group with basis $\{aC \mid a \in A\}$. It is trivial that $F[A]/C$ is abelian and generated by $\{aC \mid a \in A\}$, but how can I prove that $\{aC \mid a \in A\}$ is linearly independent?

$\endgroup$
1
1
$\begingroup$

You can avoid having to prove that by simply showing that $F[A]/C$ and $\{aC\mid a\in A\}$ have the relevant universal property... That is done in egreg’s answer in the question you link to.

To show they are linearly independent, suppose that $a_1,\ldots,a_n\in A$ be pairwise distinct and $\beta_1,\ldots,\beta_n\in\mathbb{Z}$ are such that $$a_1^{\beta_1}\cdots a_n^{\beta^n}C=eC.$$ Then $a_1^{\beta_1}\cdots a_n^{\beta^n}\in C$.

But in elements of $C$, the sum of the exponents of each $a\in A$ is $0$. This follows by noting that the map from $F[A]$ to $\mathbb{Z}$ obtained by sending $a$ to $1$ and all other elements of $A$ to $0$ must have $C$ in the kernel (since $\mathbb{Z}$ is abelian), and so if $g\in C$ then $g\mapsto 0$. In particular, the sum of the exponents of $a$ in $g$ add up to $0$. This holds for each $a\in A$.

Since $a_1^{\beta_1}\cdots a_n^{\beta_n}\in C$, then $\beta_i=0$ for each $i$. Thus, the set $\{aC\mid a\in A\}$ is linearly independent in $F[A]/C$, as claimed.

$\endgroup$

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.