Solvable Group of Polynomial of Degree 4

Question

I am studying the insolvability of polynomial of degree $>4$, i.e., indeterminates more than $4$ in the the article Galois Theory for Beginners by John Stillwell. The complete proof is given below.

If $E$ is a radical extension of $\mathbb{Q}(a_0, . . ., a_{n-1})$ for the polynomial $x^n + a_{n-1}x^{n-1} + \cdots +a_1x + a_0 =0$, then there is a radical extension $\bar E \supset E$ such that the Galois group $G_0=\text{Gal}(E/\mathbb{Q}(a_0, . . ., a_{n-1}))$ includes automorphisms $\sigma$ extending all permutations of $x_1,..., x_n$.

In simple words, if $n=4,$ $G_0$ has all permutations of symmetric group $S_4,$ on roots $x_1, x_2, x_3, x_4$ including a 3-cycle permutation $(x_1,x_2,x_3).$ But according to the the argument for $n=5$ given in the article, $(x_1,x_2,x_3)$ will be in each $G_i$ for all $i$, note here, $G_i$ normal in $G_{i-1}$ and $G_{i-1}/G_i$ abelian.

So, it appears that $S_4$ is not a solvable group, just like $S_5!!$

Obviously I misunderstood something, but can't point it out, so please provide an elaborate answer with explicit example/work-out example where I made the mistake.

The Proof:

Postscript:

I would like to be sure about the phrase "extending" of automorphisms $\sigma$ in the paper. My understanding is, any permutation of $x_1,..., x_n$ extends means permutation that work on $x_1,..., x_n$ also works on a function of $x_1,..., x_n$, is it correct? Correct me if I am wrong.

Edit:

Please provide illustration using explicit example of subgroups and their group elements.

Answer 1

Let me give you the proof that $S_4$ is solvable: $S_4 \trianglerighteq A_4 \trianglerighteq V_4 \trianglerighteq 1$ is a normal series with abelian quotients (of order 2,3 and 4 respectively). If you now look at the argument for $S_5$ with this specific normal series in mind, can you find the place were it goes wrong for $S_4$ ? Specifically: You can convince yourself that any commutator of 3-cycles is in the Klein four group and in particular, the 3-cycles themselves are not commutators.

Answer 2

$S_4$ has subgroup series $S_4 \trianglerighteq A_4 \trianglerighteq V_4 \trianglerighteq 1$, it is a normal series with abelian quotients (of order 2,3 and 4 respectively) where - $A_4 = \{e, (12)(34), (13)(24), (14)(23), (123), (124), (132), (134), (142), (143), (234), (243)\}$.

$V_4 = \{e, (12)(34), (13)(24), (14)(23)\}$, here $e$ is the identity element.

Note the following fact:

For any 2 elements $\sigma, \tau \in G_{i-1},$ the element $\sigma^{-1} \tau^{-1}\sigma\tau$ must be an element of $G_i.$

So, according to the fact, any 2 elements $\sigma, \tau \in A_4,$ the element $\sigma^{-1} \tau^{-1}\sigma\tau$ must be an element of $V_4.$ You can check it for yourself, for example, $(1 2 3), (1 2 4) \in A_4$, and we see that, $ (1 2 3)^{-1} (1 2 4) ^{-1} (1 2 3) (1 2 4)= (12)(34) \in V_4$. Another example is, $(1 3) (2 4), (1 2) (3 4) \in A_4$, and we find that $((1 3) (2 4))^{-1} ((1 2) (3 4))^{-1} (1 3) (2 4) (1 2) (3 4)= e \in V_4.$ So, the element $\sigma^{-1} \tau^{-1}\sigma\tau$ of $G_i$ could be either identity element $e$ or non-identity element of $G_{i-1}$.

Now, consider the case of $S_5$, the elements are (from [1]) -

$S_5$ has twenty 3-cycles (permutations) as elements, consider two 3-cycle $(3 5 2) , (4 1 3)$. Let's assume, $S_5$ has a normal series with abelian quotients, then the first normal subgroup, say $G_1$, must have $(3 5 2)^{-1}(4 1 3)^{-1}(3 5 2) (4 1 3)=(1 2 3)$, similarly, we can show that all twenty 3-cycles of $S_5$ should be in $G_1$.

Similarly, if $G_1$ has normal subgroup $G_2$ with abelian quotients $G_1/G_2$, then we can show, all twenty 3-cycles of $S_5$ should be in $G_2$, if we keep using the same argument, we realize that all subgroups in the normal series of $S_5$ with abelian quotients, will have all twenty 3-cycles of $S_5$, even the last group, but according to the property of "solvability" given at the end of theorem 2 on page 26, the last group of the normal series of $S_5$ is supposed to have only the identity element, thus we find that $S_5$ can not have a normal series with abelian quotients. Thus, $S_5$ is not solvable.

Conclusion:

For a polynomials of five indeterminate, there might be a case that $\text{Gal}(F(\alpha_1, \cdots \alpha_5)/F) \cong S_5$, and according to the theorem 2 on page 25, $\text{Gal}(F(\alpha_1, \cdots \alpha_5)/F)$ must have a normal series with abelian quotients, so when $\text{Gal}(F(\alpha_1, \cdots \alpha_5)/F) \cong S_5$, $\text{Gal}(F(\alpha_1, \cdots \alpha_5)/F)$ can not have a normal series with abelian quotients since $S_5$ can not have a normal series with abelian quotients.

Reference:

1. Counting the Subgroups of the One-Headed Group $S_5$ up to Automorphism by D. Samaila