
[Aug. 14th, 201511:41 am]
Jason

(Nicer version to look at)
There's something about the classical connection between group theory and roots of polynomials that I've never seen clearly explained as a thinguntoitself without a lot of surrounding material about fields and extensions and so forth.
Let's just assume you know about polynomials, you know about the complex numbers and the fundamental theorem of algebra, and you know what groups are and maybe you've heard of normal subgroups but don't see what the big deal is about them.
So I've got a polynomial with n roots, it's gotta be $(x  r1)(x  r2) ... (x  rn)$ for some roots. But that's not the form I have it in; I'm just given the coefficients in front of each power of x. Let's be concrete and think about the cases of n=2 and n=3.
$$(x  p)(x  q) = x^2  (p + q)x + pq$$ $$(x  p)(x  q)(x  r) = x^3  (p + q + r)x^2 + (pq + qr + rp)x  pqr$$
What I know in the $n = 2$ case is $p + q$ and $pq$, and what I want to find is $p$ and $q$. What I know in the $n = 3$ case is $p + q + r$ and $pq + qr + rp$ and $pqr$ and what I want to find is $p, q, r$.
Can I do this by adding, subtracting, multiplying, dividing? No! Because all of the things I do know are preserved under arbitrary permutations of the roots. $ p + q + r $ is a symmetric polynomial in the sense that it is the same polynomial after I make any consistent substitution of a permutation of $pqr$ for $pqr$. I.e. $p + q + r = q + p + r = q + r + p = \cdots$ etc. But $p$ itself  the thing I want to find  is not, because $p \ne q \ne r$.
But taking roots gets me somewhere, since when I take an $n$th root, I break symmetry by exactly a factor of $n$. How does the quadratic formula work? When I solve $x^2 + Bx + C$ by computing ${B \pm \sqrt{B^2  4C} \over 2}$, what's going on in the discriminant, there? $B$ is $(p + q)$ and $C$ is $pq$ so $B^2  4C$ is $p^2 + 2pq + q^2  4pq = p^2  2pq + q^2$ and has the set $\{p  q, q  p\}$ of square roots.
This is the crux: There exists a symmetric polynomial, $p^2  2pq + q^2$ that arises as the square of a not symmetric polynomial, namely $p  q$. And once I get my hands on a nonsymmetric polynomial I at least now have a chance of doing some further algebra and getting $p$ and $q$ by themselves.
But the $n=2$ case is easy because the full symmetry group of two things is already cyclic. If I have three things the symmetry group is 6 big, but it's not ${\mathbb Z}_6$. So by taking an $n$th root, I can reduce the amount of symmetry of the group by a factor of $n$, but I'm going to have to do it in a couple of steps.
So can I maybe find something that's not **totally** asymmetric, but just less fully symmetric, whose, let's say, square is equal to some symmetric polynomial? If so, how much less symmetric would it be? If the gap between what I start with and what I end up with is related to squaring/squarerooting, it should have a symmetry group of size 6/2=3. So, maybe I'm looking for a **cyclically symmetric** polynomial, one invariant under rotations (but not all permutations) of $p,q,r$, whose square is symmetric.
Turns out there is one! If you do out all the expansion, $$(p^2 q + q^2 r + r^2 q  pq^2  qr^2  rq^2)^2$$ is symmetric, even though $p^2 q + q^2 r + r^2 q  pq^2  qr^2  rq^2$ is merely cyclically symmetric. An important thing to note is that $p^2 q + q^2 r + r^2 q  pq^2  qr^2  rq^2$ is antisymmetric  it flips sign  if you interchange any two of $p,q$. But we're squaring it, so that signflip goes away. And the invariance under rotational symmetry of $p,q,r$ that we already had, and this flipsymmetry combine to give the full symmetry group $S_3$.
The general idea is that if both $G$ and $H$ are subgroups of $S_n$, and you have $H \triangleleft G$ and $G / H$ is ${\mathbb Z}_m$, then for any $p$ that's $H$symmetric, you can take $\omega$ to be an $m^\hbox{th}$ root of unity, and pick an arbitrary element $\pi \in S_n$ of a generator of $G/H$, and then the polynomial $$\left(\sum_{r=0}^{m1} \omega^r \pi^r(p) \right)^m$$ is $G$symmetric. Why? Because any element of $G$ can be represented as $h\pi^s$ for some $h\in H$ and $s\in \{0,\ldots,m1\}$, and $$h\pi^s\sum_{r=0}^{m1} \omega^r \pi^r(p) $$ $$= \sum_{r=0}^{m1} \omega^r h\pi^s\pi^r(p) $$ $$= \sum_{r=0}^{m1} \omega^r h\pi^s\pi^r(p) $$ $$= \sum_{r=0}^{m1} \omega^r h\pi^{r+s}(p) $$ $$= \sum_{r=0}^{m1} \omega^r \pi^{r+s}h'(p) $$ (for some $h' \in H$, since $H$ is a normal subgroup of $G$) $$= \sum_{r=0}^{m1} \omega^r \pi^{r+s}(p) $$ (since $p$ is $H$symmetric) $$= \sum_{r=0}^{m1} \omega^{rs} \pi^{r}(p) $$ (because $\pi^m = 1$) $$= \omega^{s} \sum_{r=0}^{m1} \omega^{r} \pi^{r}(p) $$ and the $\omega^{s}$ vanishes when we take the $m$th power.
So it's by taking $\omega$ to be a third root of unity and considering $p + \omega q + \omega^2 r$, which has the property that $(p + \omega q + \omega^2 r)^3$ is cyclically symmetric, which means that $((p + \omega q + \omega^2 r)^3  (p + \omega r + \omega^2 q)^3)^2$ is fully symmetric. So if only we could be certain that we could construct **any** symmetric polynomial out of $p + q + r$ and $pq + qr + rp$ and $pqr$, we'd have a way of computing $p + \omega q + \omega^2 r$ via a square root and then a cube root... but it turns out we can! 

