Understanding Abel-Ruffini I'm wondering of anyone can point me towards a proof of why we can't have a quintic formula, using concepts from basic group theory. In particular, I understand that there is some connection with symmetry groups and the lack of (non-trivial) commutative quotient groups, but I don't see what this really has to do with roots. I don't even understand what symmetry groups have to do with roots of polynomials.
There is a concise explanation on this question, but it is a bit heavy on the field theory.
Can someone explain — or point me to an explanation — for why symmetry groups appear here and what they have to do with radicals?
I am an undergraduate and don't really want to spend days reading a proof; I'm really looking for an intuitive explanation of the theorem, or at least an intuition for the connection with group theory.
 A: Symmetry groups appear in roots of polynomials as follows.  Suppose you have a polynomial with rational coefficients, and one of its roots is $\sqrt{2}$.  I claim that $-\sqrt{2}$ must also be a root.  A little experimentation with some polynomials should make this claim plausible. The intuition is that if the polynomial has rational coefficients, then when you substitute $\sqrt{2}$, there will be nothing to distinguish $\sqrt{2}$ from $-\sqrt{2}$: they are both just "square roots of 2".  In that sense, $\sqrt{2}$ and $-\sqrt{2}$ are symmetrical.  Similarly, if $3 + \sqrt{2}$ is a root, then $3 - \sqrt{2}$ must also be, and so on.  Notice that rational numbers stay fixed while $\sqrt{2}$ and $-\sqrt{2}$ are exchanged.  (Of course, all of this is false if the polynomial does not have rational coefficients, e.g. if the polynomial is $x - \sqrt{2}$.)
Let's turn this around a bit.  Let $a = \sqrt{2}$ and $b = -\sqrt{2}$.  Now write down some polynomial equation with rational coefficients involving $a$ and $b$.  I claim that if you exchange $a$ and $b$ in any such equation, then the new equation is also correct.  For example, $a^2=2$ becomes $b^2=2$.  Or, $a+b=0$ becomes $b+a=0$.  If you try to write something asymmetric in $a$ and $b$, it doesn't work. Try $a-b$.  You can't complete that to a polynomial equation with rational coefficients (unless you symmetrize in some way, for example, by writing $(a-b)^2$).  So we have $S_2$, the symmetric group on 2 letters, acting on the square roots of 2.
Now let's try cube roots.  Consider the 3 cube roots of 2: $a = \sqrt[3]{2}$, $b = w\sqrt[3]{2}$, $c = w^2\sqrt[3]{2}$.  To be definite, here $\sqrt[3]{2}$ denotes the real cube root of 2, while $w = (-1 + i \sqrt{3})/2$ denotes one of the nonreal cube roots of 1 (then $w^2$ is the other one).  It's harder to see, but still it is true that any polynomial equation in $a,b,c$ with rational coefficients must be symmetrical in all 3 variables, e.g., $a+b+c=0$.  Here the group is $S_3$, the symmetric group on 3 letters.
Now what about 4th roots of 2?  Do we get $S_4$?  It turns out, we do not.  The four 4th roots of 2 are $a=\sqrt[4]{2}$, $b=i a$, $c = -a$, $d = -i a$.  Now we have an equation $a + c = 0$ which is not symmetric in all 4 variables (for example, $a+b=0$ would be false).  The only symmetries are those for which $a$ and $c$ "go together", and the same with $b$ and $d$.  More formally, it is the subgroup of $S_4$ preserving the partition $\{\{a,c\},\{b,d\}\}$.  There are 8 such elements of $S_4$, and it turns out that the group is $D_8$, the dihedral group of 8 elements, or the symmetries of a square.  If you label the vertices of a square $a,b,c,d$ in clockwise order, you can see that opposite vertices must go together in any symmetry.
What does all this have to do with Abel-Ruffini (solvability by radicals)?  It shows that the use of radicals limits the symmetries that the roots of a polynomial can have.  For example, in the last example above, we don't get all of $S_4$.  So if the roots of polynomial are "too symmetrical", then they cannot be expressed in terms of radicals.  As it turns out, if the symmetry group of a polynomial is $S_4$, then it still can be expressed with radicals (but it will be more complicated than the $\sqrt[4]{2}$ example above), but once you hit $S_5$, the roots cannot be expressed in terms of radicals.
