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The Details:

Definition 1: A class $\mathcal{G}$ of groups satisfies the Tits alternative if for any $G$ in $\mathcal{G}$ either $G$ has a free, non-abelian subgroup or $G$ has a solvable subgroup of finite index.

Example 1: The class of finitely generated linear groups satisfies the Tits alternative.

For my PhD, I plan to establish whether or not the Tits alternative holds for a certain class, $\mathcal{G}_0$ say, of cyclically presented groups. I'm new to this, however, since I'm only a few months into the PhD and I've had to change topics as, already, all of my previous topic has been subsumed by some yet-to-be-published work (as of $6^{\text{th}}$ Feb, $2018$) by someone else.

The Question(s):

What is the big picture for the Tits alternative is in terms of Combinatorial Group Theory (or wider)? What is the motivation for studying it?

Thoughts:

Classes of groups similar to $\mathcal{G}_0$ have been studied in terms of the Tits alternative before and, in some cases, it has been fully established, according to my supervisor, who said that the Tits alternative is quite an important property.

My guess is that what's going on is something similar to how exact sequences, by means of extensions, motivate the study & classification of solvable groups.

Please help :)

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    $\begingroup$ There are many reasons to study this type of questions, for instance, given a group it is natural to ask which subgroups does it have. It also provides a way of proving that a group is nonamenable. $\endgroup$ – Moishe Kohan Feb 6 '18 at 19:46
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The Tits alternative, first proven by Jacques Tits for linear groups, says the following:

If $G$ is a finitely generated subgroup of a linear group then $G$ is either virtually soluble* or contains a non-abelian free subgroup.

In geometric and combinatorial group theory, "being virtually-$\mathcal{P}$" for some property $\mathcal{P}$ is basically the same as "being $\mathcal{P}$". Now, a group cannot both have a non-abelian free subgroup and be virtually-soluble (why?). Therefore, the Tits alternative says that containing a free subgroup is the only obstruction to solubility which a finitely generated subgroup of a linear group can have.

The Tits alternative is considered a "nice", linear-like property. Therefore, showing that a class of groups satisfies the Tits alternative shows that you are dealing with a nice class of groups (and cyclically presented groups are weird! If they were also nice then this would be...nice.).

Why do we want to show linear-like properties? Well, as an example, it is a famous open question of Gromov whether or not hyperbolic groups are residually finite**. One way of showing this would be to prove that they are all linear (as linear groups are residually finite). So a first step towards proving linearity would be to prove that they satisfy the Tits alternative. And they do. Yay! On the other hand, Misha Kapovich proved that there are non-linear hyperbolic groups. So hyperbolic groups sit in a curious middle ground, along with $\operatorname{Out(F_n)}$ and mapping class groups... To add some fuel to the fire - Agol, Wise and others proved that masses of hyperbolic groups are indeed linear (basically, every hyperbolic group you've ever heard of apart from those synthetically cooked up by Kapovich to be non-linear). So hyperbolic groups are "mostly" linear, and are morally linear. Except they some are not linear. This is an (IMO, exteremly) interesting state of affairs.

*In American English, solvable.

**You may of course ask "residual finiteness: why do we care?"...

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  • $\begingroup$ I believe it is considered open whether or not mapping class groups of surfaces are linear $\endgroup$ – Paul Plummer Feb 7 '18 at 18:15
  • $\begingroup$ @Paul Sorry, I was being sloppy and didn't check this. I think you are right though - it was claimed on the arxiv in 2003, but Farb stated that it was still open in 2006. I believe Farb... I will check the primer tomorrow (unless you have a copy to hand?) $\endgroup$ – user1729 Feb 7 '18 at 18:59
  • $\begingroup$ I don't think the primer explicitly mentions it as an open problem (at least from my quick scan), but it certainly does not say it is linear, which they would mention. I get the feeling that the paper you mention isn't accepted by the community $\endgroup$ – Paul Plummer Feb 7 '18 at 21:01
  • $\begingroup$ Isn't it "contains a solvable subgroup of finite index"? $\endgroup$ – Shaun Feb 11 '18 at 14:11
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    $\begingroup$ @Shaun Gotcha! It should have read "virtually soluble or contains a non-abelian free subgroup". The rest of the post remains unaffected by this typo. (Note that virtually soluble means precisely "contains a soluble subgroup of finite index".) $\endgroup$ – user1729 Feb 11 '18 at 15:32

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