When we say a group $G$ is finitely generated we mean it can be generated by a finite number of elements but this does not exclude the possibility of being generated by an infinite number of elements of $G$. So saying $G$ is finitely generated is not saying that it is only finitely generated but rather it is to insist on the fact that it is possible to be generated with a finite number of elements, something that is important in the study of the group and also it is something that is not satisfied by all the groups, for example the group of rationals $(\mathbb Q,+)$ cannot be generated by a finite number of rational numbers, but can for example be generated by the infinite subset of rationals $\{\frac{1}{n}\;|n\in \mathbb N^*\;\}$. Also when the group is finitely generated, the number of elements is not a characteristic of the group, for example $\mathbb Z$ can be generated by one element $\{1\}$ or by two coprime integers, so we just need to know that the group is finitely generated without giving much importance to the number of the generating elements, but sometimes it seems to me that we look at the minimum number of elements that can generate the group: for example in the group of integers we give more importance to the fact that it can be generated by only one element and call it a cyclic group for that. Also in the case of an $F$-vector space we look at linear dependence between these generators induced from multiplication with scalars from the field $F$ but in a group what sort of dependence we are looking for?
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1Not sure what the question is. Amongst a collection of would-be generators we can ask if one of them can be written as a word in the others. In that way, you might hope to reduce a given set to a minimal subset. Of course, you can ask if there might be a smaller set still..made from different generators, but questions like that tend to be profoundly difficult. – lulu May 30 '16 at 19:47
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The first question is what i'm saying is correct ? – palio May 30 '16 at 19:56
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1Sure. Knowing that a group is finitely generated tells us a lot. And you are definitely correct that the size of the generating set is not a well defined property of the group. – lulu May 30 '16 at 20:10
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1Abstractly presented groups can be pretty gruesome. For example, a subgroup of a finitely generated group need not be finitely generated. See this for the standard counterexample. – lulu May 30 '16 at 20:12
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So $\mathbb Z$ having two "basis" one of size $1$ which is ${1}$ and the other of size $2$ which is ${p,q}$ ($p$ and $q$ are coprime) is a main difference from vector space in which a basis has an exact number of elements. Is my remark relevant ? – palio May 30 '16 at 20:15
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2Yes, that's a perfectly good example. – lulu May 30 '16 at 20:16
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@lulu I think there is a little "problem" in the example i gave, because the coprime couple ${2,3}$ generate $\mathbb Z$ but it is not a "basis" for $\mathbb Z$, indeed we can find integers $a$ and $b$ such that $a2+b3=0$ with $a=3$ and $b=-2$. So ${2,3}$ does not freely generate $\mathbb Z$ while ${1}$ freely generate $\mathbb Z$. Hence the size of a freely generating set of a group $G$ is constant, and here we find the similarity with finite dimensional vector spaces whose size of basis is constant. – palio May 31 '16 at 19:13
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So I think nothing is different from vector spaces, as a vector space can be generated by subsets of different sizes but it becomes constant only under the restriction that the generators are linearly independent. – palio May 31 '16 at 19:14
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1Not entirely sure what point you are making. Yes...a free group on $n$ letters is not isomorphic to a free group on $m$ letters for $n\neq m$. is that what you mean? But a finite symmetric group is generated by all its transpositions...or by one transposition and an $n$-cycle. So... – lulu May 31 '16 at 19:24
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In the answer below of @ayadi mohammed he says that both of the two generating sets: \1) a transposition (12) and a cycle $(12..n)$\ 2) all the basic transpositions $(12),(13),...,(n-1 n)$ both are free systems, is that correct !! I thought $S_n$ could not have a free generating set, being a fintie group – palio May 31 '16 at 19:36
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1I am not familiar with the definition of "free" that he is using. However, the property in question is defined in the post and it is true that both sets of generators satisfy it. I, personally would not use that word to describe that property. (Note: the fact that I am not familiar with the term does not mean that it isn't used. But I'd say it was misleading). – lulu May 31 '16 at 19:43
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1To emphasize: the definition of "free" appearing in that response is not the definition you are using. That solver doesn't mind relations like $g^2=e$ but he would object to $g_1^{-1}g_2g_1=g_3$, say. – lulu May 31 '16 at 19:45
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if you allow torsion then you get the other relation, for example if $g_1^2=e$ and $g_2^3=e$ then you get the relation that he would object $g_1^2=g_2^{3}$ – palio May 31 '16 at 19:53
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1Again, I'm not sure what you are after here. Almost everything having to do with abstractly presented groups is hopelessly difficult. Famously, even the Word Problem, i.e. just deciding whether two words in some group represent the same element, is not generally solvable. Almost always, people assume some stronger structure. – lulu May 31 '16 at 20:03
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in group theory, I think better to say, a subset S of group G is called free if any strict subset A of S we have $\langle A\rangle \varsubsetneq \langle S\rangle $
m.idaya
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1But the number of elements of a free system that generat a finitaly group is not an invariant for this group: exemple the group $ S_n$ finite with two free system an of cardinal 2 and other n-1, – m.idaya May 30 '16 at 20:20
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each of the above systems is a minimal generator system, ie all strictly subset of each of these systems does not generate the entir group.I specify systems $A_1={(12),(123\cdot\cdot\cdot n)}$ and $A_2={(i; i+1), 1\leq i\leq n-1)}$ – m.idaya May 31 '16 at 22:05