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A free group (denoted by $F(S)$) on a set $S$, called the alphabet of the free group is set of all concatenations, of the elements of $S$ and its inverses (called alphabets). In some places, I have seen a distinction between this set $S$, and the generating set of $F(S)$, i.e. a set $X$, such that $F(s)= \langle X \rangle$ is generated by $X$. How is this true? Aren't they the same? The group generated by a set, is given by all possible products of the elements of the set, and its inverses. How is this different from the alphabet $S$ of $F(S)$.

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Certainly, the alphabet is a generating set. Besides, $F(S)$ has many other generating sets, e.g., all $F(S)$. I think you are interested in those generating sets which generate $F(S)$ freely (like the alphabet). If $T$ is such a set then there is an automorphism $f:F(S)\to F(S)$ such that $f(S)=T$. Coversely, if $f:F(S)\to F(S)$ is an automorphism then $T=f(S)$ is such a generating set. So to know all generating sets you have to describe the automorphisms of the free group. You can find this description in M.Hall "The theory of groups" and other books.

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Thanks for the answer. I have asked Derek above in the comments the following questions. Please could you tell me something about it. The number of elements of S is called the rank of the free group. So, is it possible that the rank is infinite, but there exists a finite generating set generating set $<X>=F(S)$? Also vive-versa? So, the cardinality of $X$, would depend on the map $f$ that you describe in out answer. Can you tell me more about this map? Is it bijective? –  ramanujan_dirac Mar 2 '13 at 12:07
    
"So, is it possible that the rank is infinite, but there exists a finite generating set generating set <X>=F(S)? Also vive-versa?" - No, it isn't. –  Boris Novikov Mar 2 '13 at 12:42
    
@ramanujan_dirac: "Can you tell me more about this map? Is it bijective?" - About automorphisms of free groups see M.Hall "The theory of groups", theorems 7.3.2-7.3.4. - $f$ is bijective as an automorphism. –  Boris Novikov Mar 2 '13 at 15:06
    
@ramanujan_dirac: The rank of a free group is the cardinality of the smallest generating set for that group. So a free group of infinite rank does not have a finite generating set. A free group of finite rank does have infinite generating sets, because you can add 'unnecessary' generators, but then the group will not be free on the infinite generating set. –  Tara B Mar 2 '13 at 15:52
    
@TaraB: Thanks a lot! You have made everything clear. –  ramanujan_dirac Mar 2 '13 at 16:18

Your question is not very clear. What do you mean by "inverses (called alphabets)"? Of course $S$ is not the only free generating set of $F(S)$ - there are lots more!

But I think the answer to your question is that you sometimes need to distinguish between the words on the alphabet and the group elements that they represent. For example, if you have generators $x,y$, then $xy^{-1}yx$ and $xx$ are different words, but represent the same group element.

Formally, The words on the alphabet $S \cup S^{-1}$ form a free monoid $M(S \cup S^{-1})$ (semigroup with identity), where the operation is just concatenation of words, and the identity is the empty word. There is a surjective homomorphism of monoids $\theta:M(S) \to F(S)$, where $F(S)$ is the free group on $S$, in which $\theta(xx^{-1})=\theta(x^{-1}{x}) = 1$ for all $x \in S$. So $F(S)$ is generated by $\{ \theta(x) : x \in S \}$ rather than by $S$. But it is customary in practice to identify $S$ with its set of images under $\theta$.

Added later: If $S$ is a free generating set of a free group, then you can get another one by performing any of the following operations. Permute $S$ (it's often helpful to regard $S$ as an ordered set); replace any $s \in S$ by $s^{-1}$; replace any $s \in S$ by $st$ where $s \ne t \in S$. These are called Nielsen transformations, and by performing them repeatedly, you can get to any other free generating set. They also generate the automorphism group of the free group.

It's not hard to prove that the rank of a free group is an invariant of the group, so all free generating sets have the same cardinality. For finitely generated groups, the number of homomorphisms onto a group of order 2 is $2^n$, where $n$ is the rank, whereas for infinitely generated free groups, the rank is equal to the cardinality of the group itself.

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thanks for the answer. The part on monoids was very enlightening. Cud you give me an example for "S is not the only free generating set of F(S) - there are lots more!", I think this is what I am looking for. The number of elements of $S$ is called the rank of the free group. So, is it possible that the rank is infinite, but there exists a finite generating set generating set $<X>=F(S)$? Also vive-versa? –  ramanujan_dirac Mar 2 '13 at 11:38
    
@ramanujan_dirac, you may find this useful en.wikipedia.org/wiki/… –  Andreas Caranti Mar 2 '13 at 11:43
    
Yes ramanujan_dirac, the rank is only the smallest generating set. If you look at the definition for the coproduct (free product of groups, direct sum of abelian groups) as a presentation, you take the entire underlying sets of each group as the generating set of the coproduct (a disjoint union, so that when the groups have the same underlying set, changing the elements used doesn't change the coproduct). If you just combine the smallest presentations for the two groups in the same way, the coproduct will be isomorphic. –  Loki Clock Mar 2 '13 at 13:48

If you take the generating set $\{x\}$ you can start making new words, like x+x, x+x+x, and so on, that only have to obey associativity, that every time you make a new symbol w you also make a -w, and that 0+w=w. So, the underlying set of $F(\{x\})$ is all of those words that create distinct elements, not just $\{x\}$.

The thing is, alphabet is a nonstandard term, which can mean as little as "a set" across contexts, and how you've defined it here it's obviously the same as the generating set. But it's possible the term as you're seeing it elsewhere is the underlying set, or that as well as the group operation and inverse operation (in other words, the universal-algebraic signature).

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I am not talking about F({x}), I am talking about a set $X$, such that $F(S)=<X>$, the group $F(S)$ is generated by the set $X$. –  ramanujan_dirac Mar 2 '13 at 10:09
    
Can you show the place where "alphabet" is defined as you define it? –  Loki Clock Mar 2 '13 at 10:18
    
To quote from wikipedia: "In mathematics, a group $G$ is called free if there is a subset $S$ of $G$ such that any element of $G$ can be written in one and only one way as a product of finitely many elements of $S$ and their inverses (disregarding trivial variations such as $st^{−1} = su^{−1}ut^{−1})$". Dummit & Foote names the set $S$ as the alphabet of $F(S)$. –  ramanujan_dirac Mar 2 '13 at 10:30

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