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.


1 Answer 1


See Thereom 1 from "The Topological Proof of Abel–Ruffini Theorem," Henryk Żolądek (link), which I repeat below:

Theorem 1.

The groups S(n), n ≥ 5, are not solvable.


(We follow the book of J. Browkin [2]). Because the alternating group $A(n)$ is normal subgroup of $S(n)$ with two-element quotient group, it is enough to show that $A(n)$ is not solvable. But this follows from the following observation. If the cycles $\sigma = (123)$ and $\tau = (345)$ (with one common element) belong to a subgroup $H \subset A(n)$, then the elements $[\sigma, \tau] = (\sigma(3)\sigma(4)\sigma(5)) \cdot \tau^{−1} = (145) \cdot (354) = (143)$ and $[\sigma^{−1}, \tau^{−1}] = (253)$ belong to the commutator $H^{(1)}$. The latter are also cycles with one common element.

Repeating this argument we see that all the derivative groups $A(n)^{(j)}$ contain two cycles with one common element. Therefore none of them can be trivial. 􏰆

The paper uses the definition of solvable where $G$ is solvable if $G^{(r)}=\{e\}$ for some finite $r$, where $G^{(k + 1)} = \left(G^{(k)}\right)^{(1)}$ (the derivative groups), and $G^{(1)} = [G, G]$ (the commutator group).

The proof obviously doesn't work for $n < 5$ because you can't find two three-cycles in $A(n)$ with one common element in those cases.

  • $\begingroup$ I know this criterion of solvability using derivative (commutator subgroup). This is the essentially the same idea behind Arnold's proof given in youtube.com/watch?v=RhpVSV6iCko What I want to understand is not so much the proof for unsolvability of $A_{5}$, but rather the link of theorem in my post with the unsolvability / simplicity of $A_{5}$. $\endgroup$
    – Paramanand Singh
    Commented Jul 6, 2015 at 7:42

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