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EDITED: Thanks to @Mathieu for finding a bug. The group you describe is actually isomorphic to $\text{Aut}(F_r)$ (which, of course, has a well known relation to $\text{Out}(F_r)$). To see why, you …

First you need that every vertex of $T$ whose $H$-stabilizer is nontrivial is contained in the $H$-minimal subtree $T_H \subset T$, and so each such vertex is contained in $T'$. This implies that $c_T …

To expand on Mariano's example and obtain counterexamples to (2), consider the case that $F_2=K=G*H$ is a rank 2 free group. Write a long rather random element of $F_2$ using the generators $G=\langle …

ORIGINAL ANSWER, ADDRESSING A SLIGHTLY DIFFERENT QUESTION: There is a closely related poset for which greatest lower bounds and least upper bounds indeed exist. Instead of an individual free factor $A …

A good way to visualize this issue is by looking at the fractional linear action of $SL(2,\bf Z)$ on the upper half plane model of $\bf H^2$; see section 4.2 of Serre's "Trees". The kernel of the act …

Here is what I think will be an infinite index example, although I'm missing some details of proof. Take $F = \langle a,b \rangle$, identified with the fundamental group of a rose $R$ with two petals …

I ran across this question after pondering your more recent question here. Ramiro has already described examples giving a YES answer to your final question, but here is a large and naturally arising …

The fact that braid groups are automatic, proved by Thurston and recorded in the book "Word Processing in Groups" by Epstein, Cannon, Holt, Levy, Patterson, and Thurston, leads, I believe, to an algor …

Take any pair of measured laminations $\lambda,\mu$ which each fill the surface and are transverse to each other. Take sequences $\sigma_i,\tau_i$ in Teichmuller space, such that $\sigma_i$ converges …

The quotient of the free group of rank 2 by a random, long relator has cohomological dimension 2 and is not commutative.

The gap is easily fixable in the context of the paper. Let me explain the fix after first explaining the critique of Kapovich and Lustig. The first paragraph of the proof starts by choosing a leaf $ …

I would venture that this is somewhat studied when $\Lambda$ is the subgroup of inner automorphisms, and so $Aut(G)/\Lambda = Out(G)$, the outer automorphism group, and $G / Res(\Lambda) = Ab(G)$, the …

Check out "Rank of a group" on wikipedia.

To answer 1(a), if a finitely generated group $G$ has a unique inclusion minimal generating set $S$ then $G$ is finite. The reason is that if $g \not\in S$ then $g$ is a ``nongenerator'' in that it ca …

Actually, for free abelian groups something does in fact happen along the lines of what you ask. It is just that the semidirect product idea is a little bit of a red herring. In the $Z^n$ case, what …