Articles of free groups

Isomorphic Free Groups and the Axiom of Choice

When I read about free group, the proof which concerns about two free groups $F(X)$ and $F(Y)$ are isomorphic only if $\operatorname{card}(X) = \operatorname{card}(Y)$ has a sentence going as follows: $|M(X \cup X^{-1})|=|X \cup X^{-1}|=|X|$, using the axiom of choice. Can someone give me more hint about this question or some references?

Abelianization of free product is the direct sum of abelianizations

I define $\text{Ab}(G)=G/[G,G]$ where $[G,G]$ is the commutator subgroup. I want to show that $$\text{Ab}(G_1*G_2)\cong \text{Ab}(G_1)\oplus\text{Ab}(G_2)$$ This page gives a categorical proof, but I don’t know much category theory. Can someone give a purely group-theoretic proof of this (I know the universal property of abelianizations)? By the universal property, it would suffice to show that […]

On the definition of free products

I am a little confused about the definition of free products. Given a collection of groups $\{G_\alpha\}_\alpha$ in order to create their free product, I don’t understand what properties these $G_\alpha$ must have. I tell you three different circumstances that I met studying this definition. 1) In some forums and very brief notes I read […]

$F$ is a free abelian group on a set $X$ , $H \subseteq F$ is a free abelian group on $Y$, then $|Y| \leq |X|$

I am confused by the proof a proposition: $F$ is a free abelian group on a set $X$ and $H$ is a subgroup of $F$, then $H$ is free abelian on a set $Y$, where $|Y| \leq |X|.$ The proof is: Let $X$ be well-ordered in some fashion, say as $\{ x_{\alpha} | \alpha < […]

Is every normal subgroup of a finitely generated free group a normal closure of a finite set?

Let $G$ be a group, $S\subset G$ a subset, then the smallest normal subgroup of $G$ which contains $S$ is called the normal closure of $S$, and denoted by $S^G$. My question is, if $G$ is a free group of rank $n$ for $n\in\mathbb N$, then is every normal subgroup of $G$ is a normal […]

Could the concept of “finite free groups” be possible?

Is it possible to define “finite free groups” ? could that make it easier to deal with group presentations ?

Free Group generated by S is actually generated by S.

Consider the definition of free groups via the universal property: Definition. We say that the group $F$ is the free group generated by the set $S$ if there’s a map $f:S\to F$ such that whenever there is another map $g:S\to G$ from $S$ to a group $G$, there exists a unique homomorpshism $\psi:F\to G$ such […]

Question on the presentation of $(\mathbb{R}, +)$

In this question, it is shown that $(\mathbb{R}, +)$ is not a free group. But my question is: if it is not a free group, exactly what relations is it subject to? My other question is: are there sets other than $\mathbb{R}$ that generate $(\mathbb{R}, +)$?

An alternative approach to constructing the free group.

Let $A$ be a set. We wish to construct the free group $F(A)$. It seems that this (invariably?) starts out like this: Let $A’$ be a copy of $A$, and let $\mathscr A=A\cup A’$. Let $\mathscr L$ be the set of finite strings of elements of $\mathscr A$. Then it seems the usual approach does […]

Group isomorphism concerning free group generated by $3$ elements.

From Jacobson’s Basic Algebra I, page 70, Let $G$ be the group defined by the following relations in $FG^{(3)}$: $$x_2x_1=x_3x_1x_2, \qquad x_3x_1=x_1x_3,\qquad x_3x_2=x_2x_3.$$ Show that $G$ is isomorphic to the group $G’$ defined to be the set of triples of integers $(k,l,m)$ with $$(k_1,l_1,m_1)(k_2,l_2,m_2)=(k_1+k_2+l_1m_2,l_1+l_2,m_1+m_2).$$ My thoughts: I was able to show that $G’$ is generated […]