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Let F be a free group of rank n>1.

Then F and F/F' have same rank.

Please help me!

FongV
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    Since you are new, I want to give some advice about the site: To get the best possible answers, you should explain what your thoughts on the problem are so far. That way, people won't tell you things you already know, and they can write answers at an appropriate level; also, people are much more willing to help you if you show that you've tried the problem yourself. – Zev Chonoles Jul 08 '12 at 14:47
  • Thanks for your recommendation Zev Chonoles – FongV Jul 09 '12 at 15:13

2 Answers2

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Hint. Let $X$ be a basis for $F$. Prove that $F/F'$ has the universal property of the free abelian group on $X$.

Arturo Magidin
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On the same vein as Arturo's answer but with a twist: prove that if $\,\{X\}\,$ is a set of free generators of $\,F\,$, then $\,\overline X:=\{xF'\;:\;x\in X\}\,$ is a set of free generators (i.e. a basis) of the free abelian group $\,F/F'\,$ .

Hint: Show that $$\prod_{k=1}^n x_i^{m_i}F'\in F'\Longleftrightarrow m_1=m_2=...=m_n=0$$ It could help here to show first the following: for any $\,x\in X\,$ and any word $\,w\in F\,$ , we define $\,\sum_x(w):=\,$ the sum, in $\,\Bbb Z\,$, of all the powers of the letter $\,x\,$ whenever it appears in $\,w$

For example, in $\,w=yx^2y^{-1}x^{-3}yx^2\,\,,\,\,\sum_x(w)=2-3+2=1\,\,,\,\sum_y(w)=1-1+1=1\,$

Proposition: a word (element) $\,w\in F\,$ is in $\,F'\Longleftrightarrow \sum_x(w)=0\,\,,\,\forall\,x\in X$

Note: The only way to prove the $\,\Leftarrow\,$ direction above I know of requires the use of the universal property of free groups, so we're back to Arturo's answer.

DonAntonio
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