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Questions tagged [free-groups]

Should be used with the (group-theory) tag. Free groups are the free objects in the category of groups and can be classified up to isomorphism by their rank. Thus, we can talk about *the* free group of rank $n$, denoted $F_n$.

5 votes
1 answer
78 views

Epimorphism between free groups that inject on a finite subset

I asked a question on MathOverflow (https://mathoverflow.net/q/454012/513011) where the following lemma appeared: Folklore lemma: Let $S$ be a finite subset of the free group $F_n$ of finite rank $n$....
TheMathematician's user avatar
1 vote
0 answers
85 views

Construction of left adjoint

A very standard example of left adjoints would be the functor $F:Sets\to Grps$ that maps sets to their free groups is a left adjoint to $G:Grps \to Sets$ which forgets the group structure. My question ...
Ook's user avatar
  • 211
1 vote
0 answers
49 views

The group generated by two homeomorphism $f,g$ is free group?

In my manuscript, we assume that $F_2=\langle a, b\rangle$ is free group. Also, we assume that $\varphi:F_2\times X\to X$ is generated by two homeomorphism $\varphi_a=f$ and $\varphi_b=g$. Referee ask ...
user479859's user avatar
  • 1,297
4 votes
0 answers
47 views

Strong converse of Kazhdan's property (T)

In his 1972 paper Sur la cohomologie des groupes topologiques II, Guichardet proved$^\ast$ that free groups satisfy the following strong converse of property (T): The $1$-cohomology $H^1(\mathbb F_d,\...
MaoWao's user avatar
  • 15.5k
1 vote
1 answer
77 views

Epimorphism from $F_n$ to $F_{n-1}$?

$F_n$ is the free group of rank $n$. I know that it is possible to construct an Epimorphism $f: F_n \to F_{n-1}$. How is this done? (I also want to consider the case $F_2 \to \mathbb{Z}$ but maybe ...
TheMathematician's user avatar
0 votes
1 answer
77 views

Three questions about Wikipedia's definition of Van Kampen's theorem for fundamental groups

I'm studying Algebraic Topology off of Hatcher and (unfortunately as usual) I find his definition and explanation of Van Kampen's theorem to be carelessly written and hard to follow. I happen to know ...
Nate's user avatar
  • 894
3 votes
1 answer
116 views

Normalizer of Factors in an Amalgamated Free Product

I am reading through this proof that if $H$ is a non-trivial group, then the normalizer of $H$ in the free product $G:=H \ast K$ is equals $H$ (i.e., is trivial). Most of the proof seems to generalize ...
user193319's user avatar
  • 8,040
1 vote
0 answers
98 views

Rank of subgroups of free groups

I am familiar with Schreier's theorem which states that any subgroup H of a free group G is free, as well as with the formula $$ \operatorname{rank} H - 1 = |G:H|(\operatorname{rank} G - 1) $$ I have ...
Giraud's user avatar
  • 75
4 votes
1 answer
108 views

Mapping class group of surfaces, free products, and trees

Let $ \Sigma $ be a surface, possibly with boundary. Let $ MCG(\Sigma) $ denote the mapping class group. Is it true that $ MCG(\Sigma) $ has a quotient which is a nontrivial free product $ A \ast B $ ...
Ian Gershon Teixeira's user avatar
4 votes
1 answer
77 views

Are any quotients of braid groups non-trivial free products?

The braid group $ B_1 $ is trivial and the braid group $ B_2 $ is isomorphic to $ \mathbb{Z} $. The braid group $B_3$ has the property that its central quotient (i.e., $B_3 / Z(B_3)$) is isomorphic to ...
Ian Gershon Teixeira's user avatar
2 votes
1 answer
87 views

Free product of finite groups that is outside graph theory

The free product of finite groups $ A * B $ naturally acts on a biregular graph see Free Product of two finite groups. This seems like one of the only places that free products of finite groups appear ...
Ian Gershon Teixeira's user avatar
4 votes
1 answer
130 views

A group generated by two elements with uncountably many normal subgroups

Proposition. There exists a group generated by two elements with uncountably many normal subgroups. The constructive proof below appears in Geometric Group Theory: An Introduction by Clara Löh (...
stoic-santiago's user avatar
1 vote
0 answers
143 views

If $a^{p_1} b^{q_1}\ldots a^{p_n} b^{q_n} = e$ then $S = \{a,b\}$ is not a free generating set of $G = \langle S \rangle$

Let $S = \{a,b\}$ be a generating set for a group $G$. If a non-trivial word in $a, b \in S$ equals the identity $e$ of $G$, i.e., $$a^{p_1} b^{q_1} a^{p_2} b^{q_2} \ldots a^{p_n} b^{q_n} = e$$ for ...
stoic-santiago's user avatar
7 votes
0 answers
66 views

Computing $\mathrm{Fix}(\phi)$ for autormophisms $\phi$ of free groups

Let $F_A$ be the free group generated by the finite set $A$ and let $\phi\colon F_A \to F_A$ be a group-automorphism. It is known [1] that $$ \mathrm{Fix}(\phi) = \{g \in F_A : \phi(g) = g\} $$ is (...
RB1995's user avatar
  • 319
0 votes
0 answers
31 views

$\prod_p \mathbb{Z}/p\mathbb{Z}$ is not the direct sum of $\bigoplus_p \mathbb{Z}/p\mathbb{Z}$ and a torsion-free subgroup

While I was reading "Abelian Groups" by Fuchs $(2015)$, I encountered Example $1.2$ in the chapter on Mixed Groups, which stated the following: Let $p_1,p_2,\dots,p_n,\dots$ denote different ...
John123's user avatar
  • 69

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