2
$\begingroup$

Let $F$ be the free group of rank $2$, $F^{\left \langle 1 \right \rangle} = F^2=\left \langle f^2|f\in F \right \rangle$ and $F^{\left \langle n \right \rangle} = \left ( F^{\left \langle n-1 \right \rangle} \right )^2,$ $n\in \mathbb{N}.$

Does it possible to determine the order of a group $G_n = F/F^{\left \langle n \right \rangle}$ for a given number $n\in \mathbb{N}?$

It is clear, that $|G_1| = 4$, and I believe that I can prove that $|G_2| = 128$ (and I reckon that $|G_3| = 2^{136}$). But I am not shure that $|G_n|$ is finite for all $n$.

$\endgroup$
1
$\begingroup$

If $F$ is free of rank $k$, then $|F/F^2| = 2^k$. A subgroup of index $n$ in a free group of rank $2$ is free of rank $n+1$.

It follows that we have the recursive formula $|G_{n+1}| = |G_n|2^{|G_n|+1}$.

$\endgroup$
  • $\begingroup$ I am really stupid... $\endgroup$ – Tzara_T'hong Sep 7 '16 at 10:06

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.