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5 votes
3 answers
1k views

Is this a valid "easy" proof that two free groups are isomorphic if and only if their rank is the same?

I have read on different sources that it is not possible to give a simple proof that "two free groups are isomorphic if and only if they have the same rank" using only what "a student ...
agv-code's user avatar
2 votes
2 answers
212 views

Computation of Amalgamated Product $\mathbb{Z}_4 \ast_{\mathbb{Z}_2} \mathbb{Z}_6.$

I'm trying to compute a amalgamated product $\mathbb{Z}_4 \ast_{\mathbb{Z}_2} \mathbb{Z}_6$. Let $\mathbb{Z}_4= \langle a\mid a^4 =1\rangle$ and $\mathbb{Z}_6 = \langle b\mid b^6 =1\rangle $, be a ...
Horned Sphere's user avatar
4 votes
1 answer
98 views

$F_X \cong F_Y \Rightarrow |X| = |Y|$ where is the mistake in this proof

Statement. Let $F_X, F_Y$ free groups over $X,Y$ respectively. Suppose there is an isomorphism $\phi: F_X \cong F_Y$. Then $|X|= |Y|$. My proof. Let $x \in F_X$ be a word of length one, this is, $x = ...
Numox's user avatar
  • 315
3 votes
0 answers
134 views

is $F_2$ a subgroup of any other $F_n$ for $n\geq 2$

This is more to check an argument Since $F_2$ is a group generated by words of two generators, call them $\{a,b\}$ now every other $F_n$ (provided $n\geq 2$) will be all free words of more generators $...
El Ruño's user avatar
  • 746
1 vote
2 answers
130 views

$\langle a,b\mid ab=1\rangle\cong \mathbb{Z}$

I'm new to free groups and presentation of groups, and I am having some problems with some basic facts: Let $\langle a,b\mid ab=1\rangle$ be a group presentation. I want to show that $\langle a,b\mid ...
rubikman23's user avatar
2 votes
1 answer
531 views

Show that there is a unique ring homomorphism from $\mathbb{Z}$ into any ring.

$\newcommand{\Z}{\mathbb{Z}}$ For my purposes, a ring is always assumed to be a ring with identity and so because of this my working definition of a ring homomorphism requires that $\phi(1_S) = 1_R$ ...
Irving Rabin's user avatar
  • 2,673
2 votes
0 answers
132 views

Universal Property of the Free Abelian Group on $S$.

This problem is from Dummit and Foote, 6.3.11. Problem: Let $S$ be a set. The group with presentation $(S,R)$ with $R = \{[s,t] \mid s,t \in S\}$ is called the free-abelian group on $S$ denoted $A(S)$....
Irving Rabin's user avatar
  • 2,673
0 votes
1 answer
94 views

If $K(G,1)$ and $K(H,1)$ are Eilenberg-MacLane spaces, show that $K(G\ast H,1)$ is also one.

If $K(G,1)$ and $K(H,1)$ are Eilenberg-MacLane spaces, show that $K(G\ast H,1)$ is also one. Definition. A path-connected space whose fundamental group is isomorphic to a given group $G$ and which has ...
pyridoxal_trigeminus's user avatar
1 vote
1 answer
71 views

Draw a covering of $S^1 \vee S^1$ whose fundamental group is isomorphic to $\ker \Phi : F_2 \to \mathbb Z$ with $a\mapsto 2, b\mapsto 3$

$\newcommand{\Z}{\mathbb Z}$ Let $K\leq F_2 = \langle a,b\rangle$ be the kernel of the map $\Phi: F_2 \to \Z$ sending $a$ to $2$ and $b$ to $3$. Draw a cover of $S^1 \vee S^1$ whose fundamental group ...
pyridoxal_trigeminus's user avatar
0 votes
1 answer
128 views

Any injective homomorphism $F_n\to F_n$ with image of finite index is bijective ($n\geq 2$)

$\DeclareMathOperator{\im}{Im}$ Let $\Phi: F_n\to F_n$ be an injective homomorphism between free groups with $n\geq 2$ and $\im{\Phi}$ having finite index. Prove that $\Phi$ is bijective, i.e. $\im{\...
pyridoxal_trigeminus's user avatar
1 vote
0 answers
82 views

Free left $R$-module can be endowed with $(R,R)$-bimodule structure. Basis-dependent definition?

Let $R$ be a non-commutative ring. If we have a free left $R$-module $F$, we can consider a basis $\{e_i\}_{i\in I}$ in $F$ and endow $F$ with a $(R,R)$-bimodule structure in the following way: For a ...
Elías Guisado Villalgordo's user avatar
1 vote
1 answer
200 views

A group is locally free exactly when its finitely generated subgroups are free

This is Exercise 6.1.9 of Robinson's, "A Course in the Theory of Groups (Second Edition)". According to this search for "locally free finitely generated" in the group theory tag, ...
Shaun's user avatar
  • 45.7k
4 votes
1 answer
605 views

Showing that the free group of a disjoint union is isomorphic to the free product of the corresponding free groups

P. Aluffi's "Algebra: Chapter $\it 0$", exercise II.$5.8$. Still more generally, prove that $F(A\amalg B)=F(A)*F(B)$ and that $F^{ab}(A\amalg B)=F^{ab}(A)\oplus F^{ab}(B)$ for all sets $A,B$...
mrtaurho's user avatar
  • 16.2k
0 votes
1 answer
267 views

Showing this quotient group is isomorphic to $\mathbb{Z}_2\times{}\mathbb{Z}_2$

How do I show that $\mathbb{Z}_4\times{}\mathbb{Z}_2/K$ is isomorphic to $\mathbb{Z}_2\times{}\mathbb{Z}_2$, where $K=\langle(2,0)\rangle$. I'm getting confused with the details involved here, I will ...
kam's user avatar
  • 1,285
0 votes
0 answers
370 views

Products in the free group are associative

In Aluffi's Chapter 0 there is an exercise to show that the product in the concrete construction of a free group is associative by showing that the reduction of a word is independent of the order in ...
perpetuallyconfused's user avatar

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