# Understanding statement: F free of rank r $\Rightarrow$ F/F' free abelian of rank r

I found the following exercise on the 12th page of Presentations of Groups:

1. Let $F$ be free of rank $r$. Show that $F/F'$ is free abelian of rank $r$.

First of all, $F'$ is not defined. I'm assuming it stands for a general normal subgroup of $F$.

So, what seems very strange here is that $F$ is not supposed to be abelian, but if I prove that $F/F'$ has the same rank that $F$ has, then, $F/F'$ and $F$ are isomorphic, right? But how can a non-necessarily abelian group be isomorphic to an abelian one?

It seems that someone has already posted a question here related to this: Let free group, F and F/F' have same rank.

• This says that $F_r/[F_r,F_r]\cong \mathbb{Z}^r$, where $F_r$ denotes the free group of rank $r$, i.e., with $r$ generators. Of course, $F_r$ is not isomrophic to $\mathbb{Z}^r$ for $r>1$. – Dietrich Burde Jun 27 '17 at 13:46
$F'$ denotes the derived subgroup of $F$, which is just another name for the commutator subgroup $[F,F]$. Thus $F/F'$ is just the abelianization of $F$. This should help you understand the answers in the other thread, or alternatively the answers to the question Abelianization of free group is the free abelian group.
• Ok. I got it. On the other hand, it is still strange that this factor group is abelian and isomorphic to $F$. If two groups are isomorphic, they are both abelian or both non-abelian, right? – Hilder Vitor Lima Pereira Jun 27 '17 at 13:35
• Yes, but $F/F'$ and $F$ are not isomorphic. The point is that $F$ is free of rank $r$ and $F/F'$ is free abelian of rank $r$. These are two different universal properties, so it doesn't imply that they must be isomorphic. – Arnaud D. Jun 27 '17 at 13:38