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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!

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    Thanks for your recommendation Zev Chonoles2012-07-09

2 Answers 2

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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 $X$.

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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.