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This is a well-known result, but I can't find a proof of it without using topology.

Let $m\geq2$ be an integer. Then the free group of rank $2$ contains a free group of rank $m$ as a finite-index subgroup.

user26857
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Socci
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  • It's not true. A finite index subgroup of the free group of rank $n$ has rank $k(n-1)+1$ for some positive integer $k$. – Jeremy Rickard May 04 '14 at 12:50
  • OK, I'll edit the question to $n=2$. – Socci May 04 '14 at 12:53
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    You could take the kernel of any epimorphism onto a cyclic group of order $m-1$. – Derek Holt May 04 '14 at 13:31
  • See http://math.stackexchange.com/questions/306116/the-free-group-f-2-contains-f-k. – Dietrich Burde May 04 '14 at 15:47
  • @DietrichBurde: Interesting, but I don't see anything about the index there. – Socci May 04 '14 at 16:03
  • @user147752 You realize Derek's comment above already answered your quest? – DonAntonio May 04 '14 at 16:43
  • @DonAntonio Does it always exist? – Socci May 04 '14 at 16:47
  • Of course, @user147752. This follows at once from the universal property of free groups...or "directly": if the group is free on $;{x,y};$ and $;C_{m-1}=\langle c\rangle;$ , just define $;\phi(x):=c;,;;\phi(y)=1;$ and extend in the usual way to all normal (reduced) words. – DonAntonio May 04 '14 at 16:50
  • Thank you @DonAntonio, that was easy indeed. One more question: how do you see this kernel is a free group? – Socci May 04 '14 at 21:53
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    Socci: this is called Schreier's theorem: subgroups of free groups are free. – Moishe Kohan May 04 '14 at 22:35
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    @Socci, not only what studious mentions, but there's a very specific formula to calculate the rank of a (free) subgroup of finite index in a f.g. free group: it is $;k(n-1)+1;$ , where $;k=$ the index and $;n=$ the big group's rank. In the present example, we'd have $;k=m-1;,;;n=2;$ and thus the kernel is free of rank $;m;$ ... – DonAntonio May 05 '14 at 03:56

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As mentioned in the comments one can simply look at the kernel of epimorphisms to the finite cyclic group to find the groups. As an example $$f_n: F_2 \to C_n, x \mapsto \mathrm{generator},y \mapsto id$$ and note that $[F_2 : \ker f_n]=n$, and there is a well known formula $$[F_2:H]=\frac{\mathrm{rank}(H)-1}{\mathrm{rank}(F_2)-1},$$ which can be found in Combinatorial Group Theory, by Lyndon and Schupp (Prop 3.9 Chapter I) using "combinatorial methods" rather than topological. This implies for our case that $\mathrm{rank}(\ker f_n)=n+1$, so we get that there are finite index free subgroups of rank $\geq 2$.