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Questions tagged [free-groups]

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Is a free group a product of f.g subgroups of infinite index?

Let $F$ be a free group, and let $H,K \leq F$ be finitely generated subgroups of infinite index in $F$. Is it possible that for the set of products we have $HK = F$ ?
Pablo's user avatar
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8 votes
2 answers
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Is the free abstract group residually of rank d > 2?

Let $d \geq 2$ be an integer, and let $\mathcal{F}_d$ be the family of finite groups such that $G \in \mathcal{F}_d$ if and only if every subgroup of $G$ can be generated by at most $d$ elements. Is ...
Pablo's user avatar
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Is there a nontrivial profinite word which is trivial in any group with at most d generators?

Let $F$ be a free profinite group of rank $\aleph_0$, and let $d \in \mathbb{N}$. Let $N_d \lhd_c F$ be the intersection of all open normal subgroups $L \lhd_o F$ for which $F/L$ can be generated by ...
Pablo's user avatar
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2 votes
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Exhausting a free pro-p group

Recall that for a profinite group $G$ we define the subgroup rank to be $$\sup \{d(H): H \leq_c G\}$$ where $d(H)$ stands for the minimal cardinality of a set of topological generators of $H$. Let $p$...
Pablo's user avatar
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1 vote
1 answer
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Action of the pure braid group on the commutator subgroup of a free group

Let $P=P_n$ be the pure braid group on $n$ strands and $F=F_n$ the free group on $n$ generators. I'm interested in a nice description of the action of $P$ on the derived subgroup $F'$ which somehow ...
Adrien's user avatar
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1 vote
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275 views

Explicitly showing that a free group is LERF [closed]

Let $F$ be a free group on a finite set $X$, and let $M$ be a finitely generated subgroup. Marshall Hall's theorem states that $M$ is closed in the profinite topology on $F$. That is, $M$ is the ...
Pablo's user avatar
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3 votes
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A Karrass-Solitar or Ivanov-Schupp for profinite groups

Let $F$ be a nonabelian free profinite group, $H \leq_c F$ finitely generated with $[F:H] = \infty$. Must there be some $\{1\} \neq N \lhd_c F$ such that $N \cap H = \{1\}$?
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3 votes
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Marshall Hall's theorem for surface groups [closed]

Let $\Gamma_g$ be a surface group of genus $g \geq 2$, that is we have a presentation: $$\Gamma_g = \langle x_1,y_1 \dots, x_g,y_g \vert \prod_{i = 1}^g [x_i,y_i] = 1\rangle$$ Let $H \leq \Gamma_g$ ...
Pablo's user avatar
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2 votes
1 answer
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An epimorphism into a profinite group

Let $p$ be an odd prime number, $G$ a finitely generated nonabelian profinite group, $L \lhd_o G$ a pro-$p$ group with $[G : L] = 2$. Suppose that there is a continuous surjection from $G$ onto a free ...
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Bases for free pro-p groups

Let $p$ be a prime number, $F$ a free nonabelian finitely generated pro-$p$ group, $L \lhd_o F$ and $Y$ a basis for $L$ with $y \in Y$. Is there a basis $X$ for $F$ such that $y$ is in the abstract ...
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Dense free subgroups

Let $F$ be a free pro-$p$ group (for a prime number $p$) on a finite set $X$, $\Phi$ the abstract subgroup generated by $X$, $\{1\} \neq N \lhd_c F$. Is it possible that $\Phi \cap N = \{1\}$?
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A bound on the size of the center

Let $p$ be a prime number, $F$ a free pro-$p$ group, $H \leq_c F$ of infinite index. Can it be that $$\sup_{N \lhd_o F} |Z((F/N)/C_{F/N}(HN/N))| < \infty ?$$
Pablo's user avatar
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2 votes
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Finite Cohomology and free groups

Let $F$ be a finitely generated nonabelian free profinite group, $p$ a prime number, $L \lhd_o F$ with $[F : L]$ coprime to $p$, $N \lhd_c^\infty F$ contained in $L$ with $L/N$ pro-$p$, and $N \leq H \...
Pablo's user avatar
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4 votes
2 answers
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A Karrass-Solitar theorem for surface groups

Let $\Gamma_g$ be a surface group of genus $g \geq 2$. That is, there is a presentation $$\Gamma_g = \langle x_1, y_1, \dots, x_g, y_g \vert \prod_{i = 1}^{g}[x_i,y_i] = 1\rangle$$ Is there a ...
Pablo's user avatar
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5 votes
1 answer
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Bases of surface groups

Let $\Gamma_g$ be a surface group of genus $g \geq 2$. A $2g$-tuple $(x_1,y_1, \dots,x_g,y_g) \in \Gamma_g^{2g}$ will be called a Surface Basis if we have the presentation $$\Gamma_g = \langle x_1, ...
Pablo's user avatar
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2 votes
0 answers
202 views

Accessible subgroups of free groups

Let $F$ be a nonabelian free finitely generated group, and $F = G_0 \rhd G_1 \rhd G_2 \dots$ a strictly descending subnormal chain of subgroups ($G_n \lhd G_{n-1}$ for each $n \in \mathbb{N}$) each ...
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11 votes
1 answer
401 views

Union of conjugates in free groups

Let $F$ be a (finitely generated) free group, $H \leq F$ of infinite index. Is it possible that $$ \bigcup_{g \in F} gHg^{-1} = F?$$
Pablo's user avatar
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2 votes
1 answer
286 views

Can a closure make the index finite?

Let $F$ be a free finitely generated group, $H \leq F$ of infinite index. Let $c : F \rightarrow \hat{F}$ be the embedding in the profinite completion. Denote by $\tilde{F}, \tilde{H}$ the closure of $...
Pablo's user avatar
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0 votes
0 answers
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Bases and transversals

Let $F$ be a free finitely generated group, $L \leq H \leq F$ subgroups of finite index. Given bases $B$ of $F$ and $C$ of $H$, must there be a Schreier transversal $T$ for $H$ in $F$ such that the ...
Pablo's user avatar
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2 votes
1 answer
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Making a profinite group free

Let $F$ be a free profinite group, $G$ a profinite group. Suppose that the free profinite product $F \amalg G$ is a free profinite group. Must $G$ be a free profinite group? For abstract groups the ...
Pablo's user avatar
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3 votes
1 answer
739 views

Bases of free groups

Let $F$ be a free group on a finite set $X$. Let $A \subseteq X$ be a subset of $X$ contained in some $H \leq F$, a subgroup of finite index in $F$. Must there be a basis (free generating set) for $H$ ...
Pablo's user avatar
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4 votes
1 answer
455 views

Generators of Sylow subgroups

Is there a function $f : \mathbb{N} \rightarrow \mathbb{N}$ such that for each finite supersolvable group $G$, and a Sylow subgroup $S \leq G$ we have $d(S) \leq f(d(G))$? Here $d(H)$ denotes the ...
Pablo's user avatar
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5 votes
0 answers
406 views

Profinite groups, completions, and Schreier's formula

Let $G$ be a finitely generated profinite group, and $H \leq_o G$. We say that $H$ satisfies Schreier's formula in $G$ if $d(H) - 1 = (d(G)-1)[G:H]$. We say that $G$ satisfies Schreier's formula if ...
Pablo's user avatar
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4 votes
0 answers
259 views

Schreier's formula and descending chains

For a group $G$ we denote by $d(G)$ the cardinality of a smallest set of generators. A finitely generated group $G$ is said to satisfy Schreier's formula if for every subgroup $H \subseteq G$ of ...
Pablo's user avatar
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3 votes
2 answers
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(Quadratic) equation in free group?

I know almost nothing about this field, so my following question may be stupid. We know in a free group F, if $ax^{-1}a^{-1}x=1$, then $x=t^{k}, a=t^{l}$ for some t. Now consider a more general ...
guest123's user avatar
2 votes
0 answers
126 views

Free profinite completions

Let $m,n \in \mathbb{N}$. Which residually finite groups $G$ generated by $m$ elements, have the free profinite group on $n$ generators as their profinite completion?
Pablo's user avatar
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5 votes
1 answer
452 views

Normal Subgroup Growth

Let $F$ be a free group on $d$ generators. Denote by $F_{k}$ the $k$-th term in $F$'s derived series. Put $G = F/F_k$. What is the normal subgroup growth of $G$? Explicitly, for each natural number $...
Pablo's user avatar
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4 votes
2 answers
861 views

Schreier's index formula

A finitely generated group G is said to satisfy Schreier's index formula if for every subgroup H of index k in G we have: d(H) - 1 = k(d(G) - 1). For example, a finitely generated free group satisfies ...
Pablo's user avatar
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3 votes
1 answer
190 views

Intersection growth of free profinite groups

Let $F_n$ be a free profinite group of finite rank $n$ and let $V_k$ denote the intersection of all open subgroups of $F_n$ of rank at most $k$ ($k \in \mathbb{N}$). My questions are: Can I ...
user45818's user avatar
0 votes
1 answer
446 views

finite index, self-normalizing subgroup of $F_2$ [closed]

Denote $F_2=\langle a, b\rangle$ to be the free group on two generators $a, b$. Let $H\leq F_2$ to be a subgroup with finite index $n$, so $H\cong F_{n+1}$ by Nielsen–Schreier theorem, recall that $...
Jiang's user avatar
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9 votes
2 answers
3k views

Number of subgroups of a given index of a free group

Given $n,d\in \mathbb{Z}^+$, how many subgroups of index $d$ does the free group of rank $n$ have? In case $n=1$ the question is trivial, and in case $n=2, d=2$ there are 3 such subgroups. I think I ...
Grub's user avatar
  • 333
16 votes
1 answer
2k views

"Concretely" writing down elements in a free profinite group

Let $r$ be a natural number. The elements of the free group $F_r$ on $r$ generators have a nice concrete description as "words" in the $r$ generators (and their inverses). I'd like to know if there is ...
Akhil Mathew's user avatar
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6 votes
0 answers
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Dynamics of virtual automorphisms of free group

The setup is that $F$ is a free finitely generated group, $H, H'$ are subgroups of index $2$, and $\tau:H\to H'$ is an isomorphism. Denote by $B_r$ the ball around $1$ of radius $r$ in $F$, in the ...
Rostyslav Kravchenko's user avatar
4 votes
3 answers
394 views

Algebraically Independent Numbers and Affine Linear Maps

Suppose I have two lists $\alpha_1,\dots,\alpha_k$ and $\beta_1,\dots,\beta_k$ of real numbers such that all $2k$ numbers are mutually algebraically-independent over the rationals. For each $i \in \{1,...
Derrick Stolee's user avatar
2 votes
2 answers
274 views

Generating certain subgroups of free groups via primitive elements

Let $F_n$ be the free group on $n\geq 2$ generators and let $H < F_n$ be a finite index normal subgroup. Let $P\subset F$ be the subset consisting of primitive elements (an element of $F_n$ is ...
Max Smith's user avatar
9 votes
3 answers
1k views

First-order axiomatization of free groups

Is there a way to axiomatize [non-abelian] free groups in first-order logic using the language of groups (which contains the binary operation symbol $\cdot$, and the constant symbol $e$)? Is there ...
Asaf Karagila's user avatar
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2 votes
0 answers
136 views

Elements of minimal length in normal closures of elements in free groups

Let $F_n$ be a free group of rank $n$. Let $w\in F_n$ be cyclically reduced. What can be said about the element(s) of minimal length from the $\textit{ncl}(w)$ (normal closure of $w$ in $F_n$)? Under ...
Alexey Kvashchuk's user avatar
2 votes
0 answers
603 views

Stable commutator length of elements in free groups.

http://arxiv.org/pdf/math/0611889v4.pdf (page 13) In the above paper by Danny Calegari he says that the result $\text{scl}(g) \geq 1/2$ (i.e. a stable commutator length $\text{scl}(g) := \...
Alexey Kvashchuk's user avatar
5 votes
1 answer
286 views

Automorphism classes of the free group

As is well known, the conjugacy classes of the free group $F_2$ are parametrised by cyclically reduced words, up to cyclic permutation. In particular, it's easy to tell whether two elements of $F_2$ ...
Sean Eberhard's user avatar
5 votes
0 answers
737 views

Conjugacy classes of elements in free groups. One-variable equations.

First of all, wasn't sure what could be a good title for this question. If mods think of a better name, pls feel free to change it... Let $F$ be a (non-abelian) free group of finite rank, a vector $\...
Alexey Kvashchuk's user avatar
11 votes
1 answer
394 views

A question on normal closures of elements in free groups.

Let $F$ be a free group of finite rank, and $p, b \in F$, where $b$ is a root element (i.e. not a proper power). I have a case where $p^{n_k} = V_{n_k}^{-1}b^{-1} V_{n_k} \cdot U_{n_k}^{-1}b U_{n_k}$,...
Alexey Kvashchuk's user avatar
7 votes
2 answers
490 views

How to solve this one-variable equation in a free group?

Let $F_n$ be a free group, $u, v \in F_n$, where $[u,v] \neq 1 $. Trying to show that the following equation does not have a solution in $F_n$ : $$ [v,x] = [u,v] .$$ Any ideas are appreciated. I ...
Alexey's user avatar
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9 votes
1 answer
851 views

Commutators in free groups

Let $F_m$ be the free group with $m$ generators $S:=\{x_1,\dots, x_m\}$. I am interested in the following quantity $$ F(n):=\frac{|\{w\in [F_m,F_m]: \|w\|_S\leq n\}|}{|B_S(n)|} $$ where $B_S(n):=\{w\...
M.B's user avatar
  • 2,508
8 votes
1 answer
481 views

Commutator length modulo finite index subgroups

We write $cl$ for the commutator length, i.e. the least number of commutators which multiply to a given element of a group. Given an element $g$ in the commutator subgroup of the free group $G=F_2$ ...
dlbb's user avatar
  • 83
16 votes
1 answer
918 views

Explicit path in the unitary group of a $C^*$-algebra

For $G$ a discrete group, there is a canonical inclusion $g\mapsto u_g$ of $G$ into the unitary group of the reduced $C^*$-algebra $C^*_r(G)$. Denote by $[u_g]$ the class of $u_g$ in the (topological) ...
Alain Valette's user avatar
3 votes
1 answer
922 views

Direct proof that a group is Hopfian

Consider the group $G=\langle x_1,x_2,x_3|x_1^2,x_2^2,x_3^2\rangle$. Using a slightly modified version of S. Ivanov's proof here that free groups are residually finite, I can show that this group is ...
Alex Becker's user avatar
10 votes
3 answers
814 views

Non-generating sets in a free group

Let $F_n$ be the free group generated by $x_i$, for $1\leq i\leq n$. Let $a_i$ be some elements of $F_n$, also for $1\leq i\leq n$. Is there a nice way to tell when the list $\{a_i^{-1}x_ia_i\}$ does ...
Dror Bar-Natan's user avatar
19 votes
1 answer
3k views

What are the relations between conjugates and commutators?

The following algebraic structure came up when I was thinking about invariants of coloured knots. The elements are all elements of a noncommutative free group $F$, and the operations are: $a^b= b^{-...
Daniel Moskovich's user avatar
3 votes
1 answer
3k views

Normal Subgroups of Free Products

Let $G=A\ast \mathbb{Z}$ be the free product of a group $A$ and the cyclic group $\mathbb{Z}$ and suppose $K$ is a subgroup of $G$. By Kurosh Subgroup Theorem we know that $K=F\ast (\ast_{i\in I}(K\...
Sh.M1972's user avatar
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5 votes
2 answers
530 views

Primitive subwords in a free group of rank 2

I am not sure yet about what I exactly need to prove, but I guess I can formulate a rough statement similar to the following: Suppose $w\in F_2$ is a primitive word whose length is big enough. Then ...
Damiano Lupi's user avatar