This is a re-post from MSE (because I did not get the kind of answer I wanted even after offering a bounty).

At the outset I must mention that I don't have a fairly working knowledge of Galois Theory (but do have some idea of group theory in the sense that I can understand normal subgroups).

I read the proof of unsolvability of a general quintic via radicals from J P Tignol's "Galois Theory of Algebraic Equations". Here he discusses the proof by Abel and goes on to establish the following theorem (see details in my blog post):

Theorem: Let $x_{1}, x_{2}, \ldots x_{n}$ be indeterminates and let elements $a, b$ be in field $K = \mathbb{C}(x_{1}, x_{2}, \ldots, x_{n})$ such that $a = b^{p}$ for some prime number $p$. Let $n \geq 5$ and define permutations $\sigma, \tau$ such that $\sigma$ permutes $x_{1}, x_{2}, x_{3}$ cyclically and $\tau$ permutes $x_{3}, x_{4}, x_{5}$ cyclically. If $a$ in invariant under both $\sigma, \tau$ then so is $b$.

Because of the equation $a = b^{p}$ the above theorem implies that when $n \geq 5$ there are some symmetries (invariance under $\sigma, \tau$) which remain even after taking radicals. Thus starting with the elementary symmetric function $s_{1}, s_{2}, \ldots, s_{n}$ of the indeterminates $x_{i}$ the process of taking radicals will preserve the symmetries related to $\sigma, \tau$. On the other hand each one of the indeterminates $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ is changed via at least one of $\sigma$ and $\tau$. Thus the field $K$ has elements which are changed by $\sigma, \tau$ but any element of a radical extension of $F = \mathbb{C}(s_{1}, s_{2}, \ldots, s_{n})$ is invariant under $\sigma, \tau$ and hence it is not possible to get to $K$ from $F$ via radical extensions.

On the other hand most modern treatments of unsolvability of quintic base it on the simplicity of alternating group $A_{5}$. Understanding the above mentioned theorem in Tignol's book is quite easy (simple algebraic manipulation) but its not same as regards to simplicity of $A_{5}$.

Is simplicity of $A_{5}$ somehow linked with the above basic theorem? What I need is a proper link between properties of $A_{5}$ and the above theorem. What is so special about elements $\sigma = (1,2,3), \tau = (3,4,5)$ of $A_{5}$? And why don't we have such permutations when $n < 5$ whose symmetries are preserved when taking radicals.

Update: I think I need to clarify my point very clearly. The message of the theorem described above is that for $n \geq 5$ there exists a set $P$ of permutations on $n$ symbols such that $P$ includes at least one non-identity permutation and the process of root extraction preserves the symmetries induced by the permutations of $P$. For $n < 5$ root extraction preserves the symmetries induced only by the identity permutation (trivially).

My real problem is that I am unable to map this simple concept with the comparatively difficult concept of unsolvability / simplicity of $A_{5}$. There is a feeling that probably the solvability of polynomials is a much simpler concept than the solvability of groups (as least as far the general polynomials with indeterminate coefficients are concerned). At the same time this feeling is crushed by the theorem of Galois that "a polynomial is solvable by radicals if and only if its galois group is solvable" so that these concepts are at the same level of depth/difficulty.

If it helps, I checked Tignol's book and he mentions that the theorem mentioned above is by Paolo Ruffini and he gives the reference:

P. Ruffini, Opere Matematiche (3 vols), E. Bortolotti, ed., Ed. Cremonese della Casa Editrice Perrella, Roma, 1953-1954.

The result is available in pages 162-170 in vol 2.

Further Update: I am not getting the kind of answer I need. Perhaps I need to provide a context in the language of Galois Theory. Let's assume fields to be of characteristic $0$ in what follows.

Let $f(x) \in F[x]$ be a polynomial and $K$ be splitting field of $f$. Galois theory says that the galois group of $f$ (over field $F$) is the set of automorphisms of $K$ which leave $F$ fixed. Also if $L$ is another field with $F \subseteq L \subseteq K$ then galois group of $f$ over $L$ is a subgroup of its galois group over $F$. Also note that the galois group can be viewed as a subgroup of the group of permutations of roots of polynomial $f(x)$.

It is further known that when the field $L$ is obtained by adjoining all roots of an irreducible polynomial $p(x) \in F[x]$ (of degree $k$) so that $L = F(\alpha_{1}, \dots, \alpha_{k})$ and $p(\alpha_{i}) = 0$ for $i = 1, 2, \dots, k$ then the galois group $Gal(K/L)$ is a normal subgroup of $Gal(K/F)$. This case is relevant when $L$ is a radical extension of $F$.

In our case $F = \mathbb{C}(s_{1}, \dots, s_{n}), K = \mathbb{C}(x_{1}, \dots, x_{n})$ and we are directly able to find a set of permutations of $x_{i}$ which leave every element of $L$ fixed where $L$ is a radical extension of $F$. Hence if $L$ is any radical extension of $F$ with $F \subseteq L \subseteq K$ then $Gal(K/L)$ contains two permutations $\sigma, \tau$ given above. Since radical extensions correspond to a chain of reducing (getting smaller in size) normal subgroups of $S_{n}$, this shows that the series can never get to the trivial group consisting of identity.

In my opinion the theorem by Paolo Ruffini provides a very direct and easily accessible proof of unsolvability of the $S_{5}, A_{5}$ and it is so unlike the usual proofs of unsolvability of $A_{5}$. Let me know if my views are correct. Furthermore

Is the "unsolvability of universal / general polynomials of degree 5 or greater" is a much simpler result which does not require the machinery of solvable groups?

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    $\begingroup$ The earlier question, "Can the unsolvability of quintics be seen in the geometry of the icosahedron?," may help. There are some remarkably informative answers there. $\endgroup$ – Joseph O'Rourke Jul 10 '15 at 9:53
  • $\begingroup$ Just out of interest: Many Galois theory texts prove simultanneously the simplicity of $A_{n}$ for $n >4$ first, but the fact that $A_{5}$ is simple is easy to prove directly (and the fact that $A_{5}$ is not solvable is even easier, since $A_{5}$ has no non-identity normal $p$-subgroup for any prime $p$, which is almost obvious). Once $A_{5}$ is known to be simple, a pretty easy induction gives $A_{n}$ simple for $n > 5.$ $\endgroup$ – Geoff Robinson Jul 10 '15 at 10:07
  • $\begingroup$ thanks @JosephO'Rourke for the link to the interesting question. Perhaps it will provide me some answers I am looking. $\endgroup$ – Paramanand Singh Jul 10 '15 at 10:15
  • $\begingroup$ @GeoffRobinson: What is "easy" or "difficult" varies from person to person depending on his skills. My point is that the theorem by Ruffini mentioned above does not need the concept of a group itself. One just needs to know that there are things called permutations and they can be combined/composed in straightforward manner. When we look at that way it seems much more simpler than the "simplicity/unsolvability" of $A_{5}$. continued in next comment. $\endgroup$ – Paramanand Singh Jul 10 '15 at 10:19
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    $\begingroup$ What can be easier here than showing simplicity of ${\rm A}_5$? -- Just check that there is no subset of the conjugacy class sizes $1, 12, 12, 15, 20$ of ${\rm A}_5$ which includes $1$ and which sums up to a nontrivial divisor of $|{\rm A}_5| = 60$ ... . $\endgroup$ – Stefan Kohl Jul 10 '15 at 10:49

I think this is very like the "usual" proof of the unsolvability of $A_5$. I put usual in quotes because most books skip directly to proving that $A_5$ is simple, without proving unsolvability first.

Let $G$ be a finite group. Recall that the abelianization, $G^{ab}$, of $G$ is the largest abelian quotient of $G$. Set $G' = \mathrm{Ker}(G \to G^{ab})$. Repeating in this manner, we obtain the derived series $G \supseteq G' \supseteq G'' \cdots$, and $G$ is solvable if and only if this series stops at the trivial group.

To check that $A_5$ is not solvable, then, it is enough to check that $A_5^{ab}$ is trivial. How does one usually compute $G^{ab}$? One takes a list of generators $g_1$, $g_2$, ..., $g_k$ and relations $\prod_a g_{i_a}=e$ and turns them into additive generators $h_i$ and relations $\sum h_{i_a}=0$ for an abelian group. In this case, Abel takes the relations $\sigma^3=\tau^3 = e$ and $(\sigma \tau)^5=(\sigma^2 \tau)^5=e$. This shows that the abelianization of the group generated by $\sigma$ and $\tau$ is $$\frac{\mathbb{Z} s \oplus \mathbb{Z} t}{ \langle 3s,\ 3t,\ 5s+5t,\ 10s+5t \rangle }\cong \{ 0 \}.$$

To state Abel's result in modern language, he is showing the following:

Let $\sigma_1$, $\sigma_2$, ...., $\sigma_r$ be a set of permutations in $S_n$. Let $G$ be the group they generate and suppose that $G^{ab}$ is trivial. Suppose that $f$ is a polynomial such that $f^p$ is invariant for the $\sigma_i$. Then $f$ is invariant for the $\sigma_i$.

Note that this criterion is capable of proving the nonsolvability by radicals of any polynomial which is not so solvable. Let $\Gamma \subset S_n$ be the Galois group of some nonsolvable polynomial, and let the derived series of $\Gamma$ stabilize at $G$. So $G^{ab}$ is trivial but $G$ is not. Apply the above criterion to a list of generators for $G$.

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  • $\begingroup$ thanks David! You have given exactly what I needed. I will go for an accept.. $\endgroup$ – Paramanand Singh Jul 11 '15 at 3:29

Looking at your blog post with the proof of the "Theorem", what it does prove is that for each $p$ there is no non-trivial group homomorphism $A_5 \to\mathbb Z /p \cong $ the group of $p$-th roots of unity. This is equivalent to $A_5$ not being solvable.

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    $\begingroup$ It's not exactly equivalent to non-solvability, because there are non-solvable groups like $A_5 \times C_p$ that do have such homomorphisms. $\endgroup$ – Derek Holt Jul 10 '15 at 11:05
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    $\begingroup$ Of course you are right, for the equivalence one would also need to know that all smaller groups are solvable. $\endgroup$ – user75905 Jul 10 '15 at 11:22

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