24
votes
Recognizing free groups
As indicated in the comments, it's undecidable in general to take as input a finite presentation of a group and try to output whether or not this group is free or not. This is a direct consequence of ...
23
votes
Is there a useful limit or co-limit of a diagram that has only a single object?
One example: Let $X$ be a set with a group $G$ acting on it. Consider the diagram, in the category of sets, with $X$ as its only object, but with all the elements of $G$ (considered as permutations of ...
20
votes
Accepted
Is there a non-free group $G$ whose subgroups are all freely decomposable?
Yes, there's an example.
Kurosh proved that the group $G$ with presentation
$$\langle (a_n)_{n\ge 0},(b_n)_{n\ge 1}\mid a_nb_na_n^{-1}b_n^{-1}=a_{n-1},\;\forall n\ge 1\rangle$$
has the following ...
19
votes
Accepted
A result of Schützenberger on commutators and powers in free groups
The case $n=2$ (originally due to Lyndon) admits a very nice geometric argument: one notes that elements $a,b,c$ with $[a,b]=c^2$ lead to a map from the surface $\Sigma_{-1}$ of Euler characteristic -...
18
votes
Examples of locally hyperbolic groups
Many examples can be exhibited using a theorem of Gersten:
Theorem (Gersten): Let $G$ be a hyperbolic group of cohomological dimension 2. Every finitely presented subgroup $H$ of $G$ is hyperbolic.
...
17
votes
Accepted
$\operatorname{Out}(F_n)$ is not linear for $n > 3$
There is an embedding $\text{Aut}(F_{n-1}) \hookrightarrow \text{Out}(F_n)$ for any $n \ge 2$, as follows.
First one embeds $\text{Aut}(F_{n-1}) \hookrightarrow \text{Aut}(F_n)$ by extending any ...
17
votes
Accepted
Road map to learn about $\operatorname{Out}{F_n}$
Here are some assorted recommendations.
Stallings's "Topology of Finite Graphs" and Bestvina's course notes "Folding Graphs and Applications" are a great introduction to a technique that has found ...
17
votes
Accepted
Generators for the first cohomology of free groups
They are a generating set. In fact, even more is true. Recall that for a group $G$ and a $G$-module $M$, the first cohomology group $H^1(G;M)$ is the abelian group $Der(G,M)$ of derivations $G \...
16
votes
Accepted
An endomorphism of free groups
It's injective. Indeed, the image is free of rank $k\le n$, and is injective if and only if $k=n$ (as $F_n$ is Hopfian: is not a proper quotient of itself). Since the image surjects onto the ...
16
votes
Examples of locally hyperbolic groups
Finite groups.
...
...
Fundamental groups of closed connected hyperbolic three-manifolds.
Here is a proof of the latter using far too many tools from kleinian groups. Fundamental groups of closed ...
16
votes
Accepted
Can automorphism equivalence in a free group be detected in a nilpotent quotient?
This is extremely false due to the following basic fact about nilpotent groups:
Theorem: Let $N$ be a finitely generated nilpotent group and let $f\colon N \rightarrow N$ be a homomorphism. Then $f$ ...
15
votes
Examples of locally hyperbolic groups
A locally hyperbolic group is in particular coherent (i.e. locally finitely presented), which is already a special property. To add to the examples already given by Sam Nead: ascending HNN extensions ...
13
votes
Accepted
Morse theory on outer space via the lengths of finitely many conjugacy classes
You don't misunderstand, it's a subtle point that I'm sure I'll get wrong here too. You might find the proof of a slightly more general statement in Krstić and Vogtmann's "Equivariant Outer Space ...
12
votes
Accepted
Normal closures of finitely generated subgroups of a free group
This is true, though the proof uses some heavy machinery! Theorem A.1 of Agol--Groves--Manning's appendix to Agol's proof of the Virtual Haken conjecture states:
Let $G$ be a hyperbolic group, let ...
11
votes
A result of Schützenberger on commutators and powers in free groups
(a) On the free 2-step nilpotent group $L$ on $(u,v)$, the commutator $z=[u,v]$ is not a proper power.
Indeed, since $L/[L,L]$ is torsion-free, any root of $z$ should lie in $[L,L]$ and the latter is ...
11
votes
Accepted
Database subgroups of free group
I think the answer is no. There exists a Magma command $\mathtt{LowIndexNormalSubgroups}$ that does what you want, and it does indeed find generators for each of the subgroups. I believe that a ...
11
votes
Accepted
If $K\rtimes \mathbb{Z}$ is a finitely generated group but $K$ isn't, must the fixed points of $1_\mathbb{Z}$ be a finitely generated group?
No. Fix $p\ge 2$. Take the group
$$G=\{M(x,y,z;n):(x,y,z)\in\mathbf{Z}[1/p],n\in\mathbf{Z}\}$$where $$M(x,y,z;n)=\begin{pmatrix}1 & x & z \\ 0 & p^n & y\\ 0 & 0 & 1\end{pmatrix}...
10
votes
Accepted
Maximal power in a sequence of iterated commutators in the rank two free group
A word in $F_2$ can be represented by a path on the unit square grid on the plane. Now, $w_0$ is a horizontal unit interval, $w_1$ is a vertical unit interval, $w_2$ is a unit square and the image of $...
10
votes
Accepted
A question on the fundamental group of a compact orientable surface of genus >1
Probably the easiest way to see that the map $\psi\colon H_2(G) \rightarrow H_2(G^{\text{ab}})$ is injective is as follows. Since we're dealing with a surface group, the surface $\Sigma_g$ itself is ...
10
votes
Accepted
Primitive elements in a free group with trivial projection
No, there cannot be a primitive element $w \in \ker \pi$ that is not conjugate to $x_0$ or $x_0^{-1}$.
The map $\pi$ factors through $F[x_0, x_1, \dots, x_n] / \langle \langle w \rangle \rangle$, ...
10
votes
Indicability of $\mathrm{Out}(F_n)$
Nielsen gave a finite presentation of $\operatorname{Aut}(F_n)$ 100 years ago and from that you can compute that the abelianization of $\operatorname{Aut}(F_3)$ is order 2, so no, $\operatorname{Out}(...
9
votes
Accepted
Equations in free groups satisfying all elements
Yes. Denote by $a,b$ the first two free generators of $F$.
The case $w\in F\cup\langle x\rangle$ is clear. So we can suppose, after conjugation, that $w=u_1x^{n_1}\dots u_kx^{n_k}$ with $k\ge 1$, $...
9
votes
Accepted
Equivalence of surjections from a surface group to a free group
This is true, and it is written up in lemma 2.2 of "The co-rank conjecture for 3--manifold groups" by C. Leininger and A. Reid https://arxiv.org/abs/math/0202261. They state the result in slightly ...
9
votes
Accepted
Nielsen-Schreier with operations
Let $G$ be a two element group; the free group on two generators $x,y$ with the action of $G$ interchanging them is a free $G$-group (on one generator). Its subgroup generated by $xy^{-1}$ is closed ...
9
votes
Accepted
Conjugating generators in free groups
1. About injectivity: Yves already answered about injectivity in the comments. Below is an alternative argument which works more generally for free products:
Proposition 1: Let $G=A_1 \ast \cdots \...
9
votes
Free groups are CT-groups
(1) A free group admits a free action on a tree. (Namely its Cayley graph with respect to a free generating subset).
(2) If a group acts freely on a nonempty tree, then it's a CT-group.
(Indeed, if $...
9
votes
Accepted
Finite presentability of semi-direct product of free group and its commutator subgroup
This group is not finitely presentable. Indeed, write $F$ for the given free group and $F'$ for its derived subgroup. The map
$$F\ltimes F'\to F\times F,\quad (f,g)\mapsto (f,fg)$$
is an injective ...
9
votes
Accepted
Elements of a free group that can't be inverted by automorphisms
Proposition 8.7 of https://arxiv.org/pdf/math/0303386.pdf#page107 says that the nonreversible elements are exponentially generic. This means you should get your limit in 5 with fast convergence. Note ...
9
votes
Accepted
Stable equivalence of generating sets of a finitely-generated group?
The answer is $K=2k$, which is an exercise. Here is a sketch: Let $f\colon \langle x_i\rangle\to G$ be a surjective homomorphism, and $h\colon \langle x_i\rangle*\langle y\rangle\to G$ the map defined ...
9
votes
Accepted
Shortest almost trivial element of free group
Repeating from the comments section:
This (natural and beautiful) question was previously asked and answered on this site. See Collapsible group words. It also appeared recently on math.se.
The ...
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