All Questions
Tagged with free-groups finite-groups
16
questions
3
votes
1
answer
58
views
"Almost Retractible" Abelianizations of Groups
I have two related questions.
Is there a name for a nonabelian group $G$ whose abelianization is $\bigoplus_{i=1}^n \mathbb{Z}/p_i\mathbb{Z}$ such that for each $i$ there is an element $g$ whose ...
0
votes
0
answers
49
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a typo about free groups in Dummit's Abstract Algebra
I am not sure that if there is a typo in Dummit's Abstract Algebra on page219 : Let $S=G$ and the map $\pi:F(S)\to G$ is the homomorphism extending the identity map of $S$ . the first paragragh writes ...
- 351
0
votes
0
answers
34
views
Question about groups where generators are interchangable
Let $G$ be a non-abelian finite group of rank 2 defined from generators $A$ and $B$.
For any element of $G$ defined by a word $W$, let the dual of that element be defined by a word $\overline{W}$, ...
- 1,860
4
votes
1
answer
179
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Finite extensions of finitely generated free groups. [closed]
Let $G$ be a finite group and $\phi : G \to{\rm Aut}(F_2)$ be a homomorphism. For example, we have $\phi : \mathbb{Z}/2 \to{\rm Aut}(F_2)$ that switches the generators. What can I do to see if the ...
- 193
1
vote
1
answer
72
views
Representing a group as a quotient of a free group
Consider $G=F \rtimes T$, where $F=\mathbb{Z}_3 \times \mathbb{Z}_3$ and $T=\mathbb{Z}_5$. Let $\phi : \mathbb{Z}_5 \rightarrow Aut(\mathbb{Z}_3 \times \mathbb{Z}_3)$.
It is said that any group is ...
- 43
2
votes
1
answer
136
views
Injective homomorphisms between group presentations
I'm studying the proof that $S_n = \langle s_1,...,s_{n - 1} \mid s^2_i = 1, (s_is_{i + 1})^3, s_is_j = s_js_i$ for $|i - j| > 1 \rangle$. The key part of the proof is that the subgroup of $\Gamma_{...
- 4,528
0
votes
1
answer
75
views
Does $\mathbb Z/p\mathbb Z$ a free abelian pro-p group?
As far as I know, $\mathbb Z/p\mathbb Z$ ($p$ is prime) is cyclic and so it's abelian.
It is obviously a p-group, hence it is pro-p.
And it is free, for its generator, $\langle1\rangle$, has no ...
- 549
2
votes
1
answer
167
views
Basis for a free abelian group of finite rank
My question is as follows:
Given a free abelian group $G$ of finite rank $n$, is it true that every linearly independent set of $n$ members of $G$ form a basis for $G$ (when $G$ is viewed as a $\...
- 23
4
votes
1
answer
257
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Let $G= G_1 * G_2$, where $G_1$ and $G_2$ are cyclic of orders $m$ and $n$. Then $m$ and $n$ are uniquely determined by $G$.
I'm having trouble understanding the following problem from Munkres' Topology. I have shown (a) and (b) below, for (b) I got $k=\max(m,n)$, but I don't know what I need to prove for (c). In fact, what ...
- 5,601
2
votes
1
answer
496
views
Is every finite group finitely presented? (From Michael Artin's book)
The question is:
Can every finite group G be presented by a finite set of generators and a finite set of relations?
I think the answer is yes, but I can't find a general way to decompose every ...
- 351
3
votes
1
answer
4k
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Homomorphism of Free Groups
I am reading the theorem of homomorphism of free group from Fraleigh's text in $\S$36 and could only get a fuzzy idea at best:
Let $G$ be generated by $A = \{a_i \mid i \in I \}$ and let $G'$ be ...
- 3,547
0
votes
1
answer
209
views
Surjections from free groups
I am stuck on the following:
How do I go about finding surjections from the free group of rank 2 $\mathbb{F}_2 = \mathbb{ Z∗Z}$, to the finite group of two elements $\mathbb{Z}_2$.
Also, how would ...
- 1,118
2
votes
1
answer
694
views
Commuting Elements in a Free Product of Cyclic Groups
In the free group with two generators $F_2\cong\mathbb Z *\mathbb Z$ ($*$ denotes the free product),
if two elements $a$ and $b$ commute,
then there exists an element $w\in F_2$ such that $\langle a,b\...
- 4,020
2
votes
2
answers
104
views
Subgroup of a (free) group.
Let $F$ be a group, generated by $x_1,...,x_m$ and $H$ be its subgroup such that $|F:H|=n < \infty$. How to prove that $H$ can be generated by $n(m-1)+1$ elements?
- 241
2
votes
2
answers
911
views
Let $A$ be a finitely generated abelian group. Show that $\operatorname{Hom}(A,Z)$ is a free abelian group.
My question is
Let $A$ be a finitely generated abelian group. The structure theorem says that $A$ is isomorphic to $F \times T$, where $F$ is isomorphic $\mathbb Z^m$, some $m \geq 0$, and $T $ is ...
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