# About an exercise in Serre's “Trees”

Following section 1.4, which is entitled "Constructions Using Amalgams" is the innocent-enough looking exercise:

Show that the group defined by the presentation (presumably on the generators $x_1, x_2, x_3$)

$x_2x_1x_2^{-1}=x_1^2$

$x_3x_2x_3^{-1} = x_2^2$

$x_1x_3x_1^{-1} = x_3^2$

is trivial.

Now, if one simply (well, not simply in practice but simply as in I could write code to do it much more easily than I can actually do it myself...one quickly loses track of inverses etc. and discovers new magical relations that are, in fact, too good to be true) manipulates relations, writes certain words and then tries to move generators past each other, one can obtain this result. Which leaves me to wonder: why precisely is this exercise here?

Guess 1: This exercise stands in contrast to an example in this section where we have the same set of relations except on 4 generators, so 4 relations, and so the same kind of proof does not work out at all (you can no longer move all the generators past one-another); in fact Serre uses amalgams to demonstrate that the aforementioned group contains the free group on two generators. One may perhaps be tempted (as I was, indeed, for way too long) to attempt to use amalgams to solve the exercise which is the topic of this post, to no avail. Maybe the point here is that sometimes amalgams get you nothing; that some types of questions lend themselves to proof by generator and relation manipulation but others to more "abstract" diagram chasing?

Guess 2: There is an elegant proof somehow using amalgams and it escaped me.

Guess 3: Some other reason.

Anyway, perhaps from my years of secondary, college, and finally graduate math textbook experience, I feel somehow uneasy solving an exercise at the end of a section without employing any tools at all from that section. I worry I've missed the point.

Any ideas? (Excluding perhaps that I am overthinking this and need to let it go?)

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I seem to remember proving it by hand without too much trouble by using the Hall-Witt identity, I can try to dig it up or recreate it if you like. –  ndkrempel Jan 12 '11 at 1:57
Ah, you're absolutely right! There was a moment of discussion about this when someone mentioned that we needed "a Jacobi identity for groups" but no one had heard of that, and without much faith in its existence, that avenue was abandoned for the more rudimentary one with generators and relations. This is indeed the missing link that I was looking for, as it makes the exercise a whole lot easier and more immediate (indeed, more consistent with the rest of the exercises when it comes to level of difficulty). Thanks! –  Sarah Rich Jan 12 '11 at 4:21
In case anyone is interested, I've posted an algebraic proof that this group is trivial on Math Stack Exchange. –  Jim Belk Jul 3 at 17:23