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 $n$.


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 $|G_{n+1}| = |G_n|2^{|G_n|+1}$.

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

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