Is the ping-pong lemma a difficult characterization of free groups? Or is it just me? Does someone have a nice intuition about its idea or should I carry on staring at the statement?

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    $\begingroup$ Clara Löh has a set of very nicely written lecture notes on geometric group theory where the ping pong lemma is explained in its easiest form (page 69ff). Pierre de la Harpe's Topics in Geometric group theory is also a place to look at for such questions. The ping pong lemma is called Table-Tennis lemma there and can be found with several applications on page 25ff. $\endgroup$
    – t.b.
    Aug 26, 2012 at 20:41
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    $\begingroup$ A very gentle, helpful (for me, at least) description is given in "Groups, Graphs and Trees" by Meier, beginning on Page 64. $\endgroup$ Aug 27, 2012 at 13:49
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    $\begingroup$ I second the recommendation of @unit3000-21; Meier's book is an excellent source for geometric group theory when you're starting out. But I must say Clara Löh's book is gorgeous! $\endgroup$
    – user641
    Aug 27, 2012 at 19:09
  • $\begingroup$ Aren't the name of the ping-pong lemma and its proof self-explanatory? $\endgroup$ May 23, 2015 at 23:25

2 Answers 2


I'm an outsider, so maybe what I'm saying is silly. I would just think about the Tits alternative application since that really gives all the intuition. To apply it---and we'll just look at the complex case---you should produce two matrices $A$ and $B$ that have the following properties. $A$ has a dominant eigenvalue $\lambda$ (i.e., the eigenspace of $A-\lambda I$ is one-dimensional and all other eigenvalues of $A$ have modulus strictly less than $|\lambda|$) with corresponding eigenvector $v$ and $A^{-1}$ has a dominant eigenvalue $\mu$ with coresponding eigenvector $w$. Similarly, let's suppose that $B$ and $B^{-1}$ have these dominant eigenvectors with corresponding eigenvectors $v'$ and $w'$.

Then if we consider $v,w,v',w'$ as being points in projective space then the dominant eigenvalue hypotheses say that there are open disjoint neighbourhoods of $v,w,v',w$, say $U_v, U_w, U_{v'}, U_{w'}$ such that there is some $N$ such that for $n\ge N$ we have: $A^n$ maps each of these neighbourhoods into $U_v$, $A^{-n}$ maps all these neighbourhoods into $U_{w}$, $B^n$ maps each of these neighbourhoods into $U_{v'}$, and $B^{-n}$ maps all these neighbourhoods into $U_{w'}$.

Now I think of it as being like a finite-state machine with four states labelled $U_v, U_w, U_{v'}, U_{w'}$ and an alphabet $A^N, A^{-N}, B^N, B^{-N}$ with transitions as described before; e.g., if you are in state $U_w$ and you read $A^N$ then you transition to state $U_v$ (and we'll read words right-to-left).

Now you can see that the group generated by $A^N$ and $B^N$ is free since if words $W$ and $W'$ on $A^N, A^{-N}, B^N, B^{-N}$ are the same then when we let $W$ act on $U_v$ it will send it to $U_v$ if and only if the first letter of $W$ is $A^N$; to $U_w$ iff the first letter of $W$ is $A^{-N}$, etc. Thus we see that if $W=W'$ then the first letters of $W$ and $W'$ must be the same. Now we cancel and keep going. Well, maybe this isn't what you wanted, but I tried my best and maybe we're both better off because of this. Palin out.

  • $\begingroup$ What makes you call yourself an 'outsider'? $\endgroup$
    – Tara B
    Feb 18, 2013 at 15:35
  • $\begingroup$ I just meant that I do not really work with groups, free groups, etc., although I do find the Tits alternative to be rather bewitching. $\endgroup$ Feb 18, 2013 at 15:47
  • $\begingroup$ Ah, I see. Sometimes it's perhaps easier to explain something that isn't quite in your field (but that you nevertheless understand), because you will have learnt it more recently, or had to think about it more to feel you understand it, or something like that. $\endgroup$
    – Tara B
    Feb 18, 2013 at 15:52
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    $\begingroup$ More of a maverick than an outsider, really. $\endgroup$ Feb 18, 2013 at 16:38

The basic idea is very simple and nice. Suppose that $A$ and $B$ are matrices, or more generally maps, such that $Av$ and $Bw$ are different for all $v$ and $w$. The intuition is that $A$ maps in one direction and $B$ in the other ($v$ and $w$ being balls crossing the "ping-pong table").

Now suppose you have two words in $A$ and $B$ that are equal, say $$ ABBABA=AAABBB. $$ Let's multiply by $A^{-1}$ and $B^{-1}$ as needed to strip off any leading identical letters; we get $$ BBABA=AABBB. $$ Well, then with $w=BABAx$ and $v=ABBBx$ (for any ping-pong ball $x$) we get $Av=Bw$, which was supposed to be impossible. Done!


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