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Can someone help me with the following question?

Let $F=\langle x_1,\ldots,x_n \rangle $ be a finitely-generated free group . Let $p$ be a fixed prime, and $H$ be the normal subgroup of $F$ defined by :$H= F^p [F,F]$ .

Prove that $|F:H| = p^n $ .

Hope someone will help me!

Thanks!

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1  
Well $H$ is a normal subgroup of $F$ and $F/H$ is abelian (because $[F,F] < H$) and $F/H$ has exponent $p$ (because $F^p < H$). So you should be able to show that $|F/H| \le p^n$. –  Derek Holt Dec 27 '12 at 21:09
    
@DerekHolt: $F/H$ is obviously elementary abelian. It is generated by at most $n$ elements, so its order is at most $p^n$ ... My problem is with the other direction. Why does it follow that $|F:H| $ is exactly $p^n$ ? Can you help me figure it out? thanks! –  katya Dec 27 '12 at 21:34
    
@katya: One exclamation point is enough - I have edited your comment. –  Zev Chonoles Dec 28 '12 at 0:46
1  
@katya: It would be helpful if you said, when you asked the question, that you could prove $|F/H| \le p^n$ and needed help proving $|F/H| \ge p^n$. The easiest way to prove that is to observe that the elementary abelian group $G$ of order $p^n$ has $n$ generators and is therefore isomorphic to a quotient $F/K$ of $F$. Since $G$ is abelian of exponent $p$, we have $F^p[F,F] \le K$ and hence $|G| = |F/K| \le |F/H|$. –  Derek Holt Dec 28 '12 at 9:22

1 Answer 1

up vote 0 down vote accepted

I have an idea: let us check the $\,n\,$ elements

$$x_1H\,,\,x_2H,\ldots,x_nH\in F/H$$

and suppose we have integers $\,e_i\in\{0,1,...,p-1\}\,\,,\,i=1,2,...,n\,$ , s.t.

$$\prod_{k=1}^nx_k^{e_k}H=H\Longleftrightarrow\prod_{k=1}^nx_k^{e_k}=xg\in H\,\,,\,x\in F^p\,\,,\,g\in F'\Longrightarrow$$

$$\Longrightarrow x^{-1}\prod_{k=1}^nx_k^{e_{k}}\in F'$$

We can put $\,x=\prod_{i=1}^rw_i^p\,$ , where the $\,w_i\,$ are (free) words in the letters $\,x_1,...,x_n\,$ . Now a definition:

Definition: If $\,w\in F\,$ , then the exponential sum $\,e_{x_i}\,$ of $\,x_i\,$ in $\,w\,$ is the sum of all the exponents of $\,x_i\,$ in $\,w\,$.

For example, if $\,w=x_1^3x_2x_4^{-3}x_1^{-2}x_6x_1\,$ ,then $\,e_{x_1}=3-2+1=2\,\,,\,e_{x_2}=1\,\,,\,e_{x_4}=-3\,$ , etc.

Thus, the exponential sum of all the generators $\,x_1,...,x_n\,$ in $\,x\,$ is a multiple of $\,p\,$, since $\,x\,$ is the product of some words $\,w_1,...,w_s\in F\,$ to the $\,p-$th power, and the same is true for $\,x^{-1}\,$ , of course.

Then, the exponential sum of each $\,x_i\,$ in $\,x^{-1}\prod_{k=1}^nx_k^{e_k}\,$ is of the form $\,t_ip+e_i\,\,,\,\,t_i\in\Bbb Z\,$ .

But since a word $\,w(x_1,...,x_r)\,$ in free generators is a commutator element iff every word's letter's exponential sum is zero, we get

$$t_ip+e_i=0\Longleftrightarrow t_i=e_i=0$$

since $\,0\leq e_i\leq p-1\,$ .

This proves the $\,n\,$ elements above are linearly independent over $\,\Bbb F_p:=\Bbb Z/p\Bbb Z\,$ and we have the other inequality $\,[F:H]\geq p^n\,$

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Great argument ! thnks ! –  katya Dec 28 '12 at 7:37
    
Any time. I actually enjoyed messing with combinatorial group theory once again, after so many years. Yours was a very nice question. –  DonAntonio Dec 28 '12 at 11:07

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