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Let $S$ be any set and let $f:S \rightarrow F$ denote the free group on $S$. By definition, this means that for any group $X$ and any function $g:S \rightarrow X$ there exists a unique homomorphism $h:F \rightarrow X$ such that $h \circ f = g$.

1) How starting from this definition I do conclude that cannot have equations of the type $x^{3}=1;~xyx^{2}=1~;~\left( xy\right) ^{3}=1$ ?

2) Intuitively, the group $F$ to be free on $S$ it means that it is vectorial space over $\mathbb{Z}$, that is, every element of $F$ is a lineal combination on $\mathbb{Z}$ of elements of $S$?If not, then the one means what $F$ to be free on $S$?

3) Can anybody given an application of the following fact?
"every group is a quotient of a free group"

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Are the equations in (1) supposed to be true for all elements of $F$ or just some? – lhf May 17 '12 at 20:27
You can consider all. – User2040 May 17 '12 at 20:35
up vote 1 down vote accepted

The statement that any group is a quotient of a free group is the start of the idea of generators and relations for groups, which gives a method of describing a group in a small or finite amount of information. So we can describe the infinite trefoil group $T$ as given by 2 generators $x,y$ and one relation $x^2=y^3$. This leads to an exact sequence

$$1 \to N \to F\{x,y\} \to T \to 1.$$

This leads on to all sorts of questions, such as can you list the elements of $T$, how to describe the elements of $N$, and so on. Very large finite groups may have relatively small descriptions in terms of generators and relations.

You will find more about question of this type in books on Combinatorial Group Theory.

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  1. To show that $F$ cannot satisfy $x^3=1$ as an identity, it suffices to find a group $G$ and an element $a\in G$ that does not satisfy the identity. Indeed, let $G$ be a group and $a\in G$ be such that $a^3\neq 1$. Define $g\colon S\to G$ be $g(s)=a$ for all $s\in S$. Then there is $h\colon F\to G$ such that $h(s)=a$ for all $s\in S$. In particular, since $h(s^3) = a^3\neq 1$, it follows that $s^3\neq 1$, so there are elements of $F$ that do not satisfy the identity $x^3=1$.

    Similarly with the other identities. In fact, the point of a "free" group is that it is "free of any identities except those that are direct consequences of the axioms" (associativity, identity, inverses). Given any nonempty reduced word, it suffices to find a single group where that word is not satisfied identically to show it cannot be an identity in the free group.

    And, of course, one such group can be a free group appropriately constructed as a set of reduced words with "concatenate-and-cancel" as the operation; you do not need to prove that this is a free group, just that it is a group in order to get the desired conclusion.

    If you wish to show that no element (other than the identity) satisfies $x^3=1$, then you can argue by showing that the subgroup generated by $x$ is free of rank $1$, and proceed as above. Similarly for the rest.

  2. I have no idea what you mean by "vectorial S-space", so it is impossible for me to answer this question. That $F$ is free on $S$ means that $F$ and $S$ satisfy the relevant universal property.

  3. There are numerous applications; e.g., the definition of the Schur multiplier of a group relies on being able to express a group as a quotient of a free group.

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Another characterization is that every element of $F$ can be written uniquely as a product of elements of $S$ and their inverses, after eliminating trivial factors of the form $xx^{-1}$ (assuming $f$ is inclusion). – lhf May 17 '12 at 20:52
I wanted to say vectorial space over $\mathbb{Z}$. – User2040 May 17 '12 at 21:21
Vectorial spaces are usually defined as modules over division rings (commutative, i.e. fields. or non-commutative). As $\,\mathbb{Z}\,$ is not a division field you can't have $\,\mathbb{Z}\,$-vector spaces. Also, and if you meant $\,\mathbb{Z}\,$-module, these are exactly the same as abelian groups, and a free group is abelian iff it is freeley generated by one single element. – DonAntonio May 17 '12 at 22:40
Thank you very much! – User2040 May 17 '12 at 23:28

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