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.
