# 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 $2$.

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.

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 $$2$$.
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 $$H$$.
• 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 $G/H$. – Achim Krause Jul 16 '20 at 18:46