Galois Theory and Number of Solutions Is the Galois group of an irreducible separable polynomial of degree $n$ isomorphic to a group on exactly $n$ letters? Is that enough to prove that a degree $n$ polynomials has $n$ roots.
What I mean by three letters is that you can write the group as permutations on three letters for example S_3 = all permutations and C_3 = just cycles are both on three letters.
 A: If by "group on exactly $n$ letters", you specifically mean the symmetric group $S_n$, then the answer is no, not in general. For example, see PlanetMath - the Galois group of an irreducible cubic polynomial over $\mathbb{Q}$ will be $A_3\cong\mathbb{Z}/3\mathbb{Z}$ when its discriminant is the square of a rational number.
Also, the Fundamental Theorem of Algebra is really a consequence of topological properties of $\mathbb{R}$ and/or $\mathbb{C}$ - as is mentioned on Wikipedia, it is often remarked that it is not quite a theorem about algebra.

Reading your comments, I'm guessing perhaps you want "$G$ is a group on $n$ letters" to be synonymous with "$G$ can be viewed as a group of permutations of $n$ objects which acts transitively". If that is the case, then the Galois group of an separable irreducible polynomial of degree $n$ is always a group on $n$ letters, because the Galois group must indeed act transitively on the roots of the polynomial. If it didn't, we could break up the set of roots into the orbits of the action of the Galois group, and this would force the polynomial to factor - for example, if $f=\prod_{i=1}^n(x-\alpha_i)$, then if the Galois group permuted $\alpha_1,\ldots,\alpha_k$ amongst themselves, and permuted $\alpha_{k+1},\ldots,\alpha_n$ amongst themselves, then we could factor $f=gh$ where $g=\prod_{i=1}^k(x-\alpha_i)$ and $h=\prod_{i=k+1}^n(x-\alpha_i)$, and $g$ and $h$ are guaranteed to be polynomials over our base field because they are fixed by the action of the Galois group.
Also note that if a group $G$ acts transitively on a set of $n$ elements, then $|G|\geq n$, which also seems to coincide with what you are wanting the group to satisfy.
A: You don't need to know anything about Galois theory to prove that a polynomial of degree $n$ over a field $F$ has at most $n$ roots. This is a straightforward consequence of the division algorithm in $F[x]$.
A: The Galois group is a permutation group on $n$ letters, i.e., a subgroup of $S_n$. See http://en.wikipedia.org/wiki/Galois_theory#The_permutation_group_approach_to_Galois_theory. BTW, you probably need to assume that the polynomial is separable, i.e., does not repeated roots.
A: Here is a nice definition of Galois group from Abhyankars book "Lectures on Algebra" -- Let $f$ be an irreducible polynomial of degree $n$ with coefficients in some field $K$ whose roots are $a_1,a_2,..a_n$ then the Galois group $G$ consists of those permutations which preserves the relations between them i.e. G is the set of all those permutations $\sigma$ of the symbols ${1,..,n}$ such that $\phi(a_{\sigma(1)},..,a_{\sigma(n)})=0$ for very one variable polynomial $\phi$ for which $\phi(a_1,..a_n)=0$
