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This is Theorem B of Swan's famous paper Groups of cohomological dimension one, in which he removes the 'finitely generated' hypothesis from Stallings' theorem. Theorem B states: Let $G$ be a tors …

There are many, many examples of large, residually finite groups with trivial outer automorphism group. Indeed, any nonabelian virtually special group is large and residually finite. Furthermore, if t …

This doesn't answer the main question, but it does address the parenthetical question of the decidability of computing presentations in matrix groups. In so doing, I hope it helps to clarify Joel and …

As Yves points out, the answer is 'no' if you replace 'malnormal' by 'almost malnormal'. However, the answer to your first question is 'yes': it is the case that the Karrass--Solitar result remains t …

I believe that Ben McReynolds thought about this for infinite-index subgroups of surface groups---you might ask him if he got anywhere. I had some reasoning similar to Ian's to suggest that it should …

Here's a short proof by hand that $K$ is characteristic in $F_2$. The natural map $Q_8\to V=\mathbb{Z}/2\oplus\mathbb{Z}/2$ induces $\eta:F_2\to H_1(F_2,\mathbb{Z}/2)\cong V$. The kernel $N=\ker\eta …

The answer to your question is 'no'. Consider any homomorphism $f:A*F\to G$ which is injective on $A$. Then $\ker f$ is an ideal in your sense (and this is necessary and sufficient). An infinite inc …

I think your problem is solved if I tell you that the group is supposed to act freely (and specially, if you like) on a CAT(0) cube complex, or equivalently to be the fundamental group of a non-positi …

No. Let $\Gamma$ be the graph with two vertices and no edges - the non-abelian free group of rank two - and let $g$ be the commutator of the two generators $s_1$ and $s_2$. Then $g$ is certainly non …

As James points out, the paper of Davis and Januskiewicz proves the inverse. To see that the answer to your question is 'no', consider the right-angled Coxeter group whose nerve graph is a pentagon. …

Yes! See my recent preprint Alternating quotients of free groups. I expect that what you want is well known, but I too couldn't find it in the literature. In fact, I prove the much stronger result t …

No. Ollivier & Wise's version of the Rips Construction gives, for any finitely presented group $Q$, a finitely presented group $G$ of cohomological dimension 2 and a surjection $G\to Q$ such that the …

Well, if your definition of a Fuchsian group is a finitely generated, discrete subgroup of $PSL_2(\mathbb{R})$, then the answer is 'no'. $H$ could be the direct product of $G$ with an arbitrary finit …

I think it's pretty clear that you also need a presentation for $N$ and to understand the map $G\to G/N$. (It's hard to say for sure, as you don't say how $G$ and $N$ are given to you.) Given that i …

Let $G=\langle x,y,c\mid w(x,y)=c\rangle$ (which, in this case, happens to be a free group). The set of solutions to the system that you are looking for corresponds precisely to the set of homomorphi …