4
$\begingroup$

I am having difficulties in solving the following problem.

Let $G$ be a group with a free subgroup of rank $2$. Let $H\leq G$ be such that $[G:H]<\infty$. Then $H$ also contains a free subgroup of rank $2$.

We know by Nielsen-Schreier theorem that a subgroup of a free group is also free. But in this problem $G$ is not necessarily free but contains a free subgroup. How to approach this problem? Any hint or idea will be highly appreciated. Thanks in anticipation.

$\endgroup$
5
$\begingroup$

Let $F$ be the free subgroup of rank $2$ in $G$.

Then $|G:H|$ finite implies that $k := |F:H \cap F|$ is also finite, and by the Nielsen-Schreier Theorem $H \cap F$ is free of rank $k+1$.

So $H \cap F$ and hence also $H$ contains a free subgroup of rank $2$.

$\endgroup$
4
$\begingroup$

Let $n=|G:H|$ be the index of $H$ in $G$, and let $a, b\in G$ be the generators of $F$. Then $\langle a^{n!}, b^{n!}\rangle$ is free of rank two, and is contained in $H$.

$\endgroup$
2
  • 3
    $\begingroup$ Slight correction: if $H$ isn't normal in $G$, it is not true that every $n$-th power $x^n$ lies in $H$. For example, consider $G=\Sigma_3$ and $H$ a cyclic subgroup of order $2$. However, any $x^{n!}$ lies in $H$, by considering the action of $x$ on $G/H$. $\endgroup$ – Achim Krause Jul 16 '20 at 18:46
  • 1
    $\begingroup$ @Achim Thanks, nice spot! $\endgroup$ – user1729 Jul 16 '20 at 19:04

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.