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.


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.


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