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I'm trying to show that a group generated by elements $x,y,z$ with a given relation $xyxz^{-2}=1$ (where $1$ is the identity) is in fact a free group.

What are some usual ways of going about this kind of problem? Just hints please!

Regards

3 Answers 3

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Are you aware of the relations of your problem to the subject of presentation of groups, in general, and in particular to the Nielsen-Schreier therorem and Nielsen transformations? This should qualify as one of the "usual ways" of going about this kind of problem at least.

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    Ye$s$, thank you. I've realized that my question was way to vague. I've edited it and put in the actual relation.2011-12-01
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You have $y=x^{-1}z^2x^{-1}$ from your relation, so $G=\langle x,z\rangle$ without any defining relation.

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(This is essentially just an expansion of Boris Novikov's solution.)

There is an important technique known as Tietze transformations for manipulating presentations of groups. The two Tietze transformations are:

  1. Add a new relation which can be derived from the existing relations. Inversely, remove a relation that can be derived from the others.

  2. Add a new generator, together with a relation that expresses it in terms of the existing generators. Inversely, you can remove a generator if this generator only appears in one relation, and this relation defines the generator in terms of other generators.

A special case of (1) is to replace a relation with an equivalent relation. Technically, this involves two applications of (1): you must first add the new relation, and then remove the original.

Tietze's theorem is that these moves do not change the isomorphism type of a group defined by a presentation. Moreover, if two finitely presented groups are isomorphic, it is possible to get from the first presentation to the second using a finite sequence of Tietze transformations.

Now, the given group has the following presentation: $ \langle x,y,z \mid xyxz^{-2} = 1\rangle. $ The given relation is equivalent to the relation $y = x^{-1}z^2 x$, so we can replace it using two Tietze transformations of type (1): $ \langle x,y,z \mid y=x^{-1}z^2 x^{-1}\rangle. $ We can now use a Tietze transformation of type (2) to remove the generator $y$. This leaves $ \langle x,z \mid -\rangle $ which is a presentation for the free group of rank two.

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    Does this mean the group is not actually generated by the three elements $x,y,z$?2017-12-05