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For two free groups with finite ranks, are they quasi-isometric to each other?

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closed as off-topic by Harish Chandra Rajpoot, user91500, Antonios-Alexandros Robotis, Shailesh, choco_addicted Mar 16 '16 at 4:56

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You can show that $F_{nm+1}$ is a subgroup of $F_{n+1}$ of index $m$, which implies that for any $n, m$ we have that $F_{nm+1}$ is quasi-isometric to $F_{n+1}$. In particular, setting $n = 1$ gives that all free groups of finite rank at least $2$ are quasi-isometric to $F_2$ and hence to each other. ($F_1 \cong \mathbb{Z}$ is not quasi-isometric to the others.)

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  • $\begingroup$ Thank you, I wonder how to show the claim you used: Every group is quasi-isometric to any subgroup of finite index? $\endgroup$ – Katherine Mar 16 '16 at 5:08
  • $\begingroup$ nice..................+1 $\endgroup$ – Bhaskara-III Mar 16 '16 at 5:23
  • $\begingroup$ @Katherine: If you have a followup question, it is better to post it as a different question. However, to avoid the question being closed, it is good to include your thoughts about the question, including what you have tried, where you got stuck, and so on. $\endgroup$ – Lee Mosher Mar 16 '16 at 13:34
  • $\begingroup$ @LeeMosher Thank you. $\endgroup$ – Katherine Mar 16 '16 at 16:42

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