Lemma 11.2.1.
If \(p\) is prime, then any subgroup of \(S_p\) that contains a transposition and a cycle of length \(p\) must be all of \(S_p\text{.}\)
Section 11.2 Insolvability of the Quintic
We are now in a position to find a fifth-degree polynomial that is not solvable by radicals. We merely need to find a polynomial whose Galois group is \(S_5\text{.}\) We begin by proving a lemma.
Proof.
Let \(G\) be a subgroup of \(S_p\) that contains a transposition \(\sigma\) and \(\tau\) a cycle of length \(p\text{.}\) We may assume that \(\sigma = (1 \, 2)\text{.}\) The order of \(\tau\) is \(p\) and \(\tau^n\) must be a cycle of length \(p\) for \(1 \leq n \lt p\text{.}\) Therefore, we may assume that \(\mu = \tau^n = (1, 2, i_3, \ldots, i_p)\) for some \(n\text{,}\) where \(1 \leq n \lt p\) (see Section 9.1). Noting that \((1 \, 2)(1, 2, i_3,\ldots, i_p) = (2, i_3, \ldots, i_p)\) and \((2,i_3, \ldots, i_p)^k(1 \, 2)(2,i_3, \ldots, i_p)^{-k} = (1 \, i_k)\text{,}\) we can obtain all the transpositions of the form \((1n)\) for \(1 \leq n \lt p\text{.}\) However, these transpositions generate all transpositions in \(S_p\text{,}\) since \((1 \, j)(1 \, i)(1 \, j) = (i \, j)\text{.}\) The transpositions generate \(S_p\text{.}\)
Example 11.2.3.
We will show that \(f(x) = x^5 - 6 x^3 - 27 x - 3 \in {\mathbb Q}[x]\) is not solvable. We claim that the Galois group of \(f(x)\) over \({\mathbb Q}\) is \(S_5\text{.}\) By Eisenstein’s Criterion, \(f(x)\) is irreducible and, therefore, must be separable. The derivative of \(f(x)\) is \(f'(x) = 5 x^4 - 18 x^2 - 27\text{;}\) hence, setting \(f'(x) = 0\) and solving, we find that the only real roots of \(f'(x)\) are
\begin{equation*}
x = \pm \sqrt{ \frac{6 \sqrt{6} + 9 }{5} }\text{.}
\end{equation*}
Therefore, \(f(x)\) can have at most one maximum and one minimum. It is easy to show that \(f(x)\) changes sign between \(-3\) and \(-2\text{,}\) between \(-2\) and \(0\text{,}\) and once again between \(0\) and \(4\) (Figure 11.2.2). Therefore, \(f(x)\) has exactly three distinct real roots. The remaining two roots of \(f(x)\) must be complex conjugates. Let \(K\) be the splitting field of \(f(x)\text{.}\) Since \(f(x)\) has five distinct roots in \(K\) and every automorphism of \(K\) fixing \({\mathbb Q}\) is determined by the way it permutes the roots of \(f(x)\text{,}\) we know that \(G(K/{\mathbb Q})\) is a subgroup of \(S_5\text{.}\) Since \(f\) is irreducible, there is an element in \(\sigma \in G(K/{\mathbb Q})\) such that \(\sigma(a) = b\) for two roots \(a\) and \(b\) of \(f(x)\text{.}\) The automorphism of \({\mathbb C}\) that takes \(a + bi \mapsto a - bi\) leaves the real roots fixed and interchanges the complex roots; consequently, \(G(K/{\mathbb Q} )\) contains a transposition. If \(\alpha\) is one of the real roots of \(f(x)\text{,}\) then \([\mathbb Q(\alpha) : \mathbb Q] = 5\) by Exercise 5.4.28. Since \(\mathbb Q(\alpha)\) is a subfield of \(K\text{,}\) it must be the case that \([K : \mathbb Q]\) is divisible by \(5\text{.}\) Since \([K : \mathbb Q] = |G(K/{\mathbb Q})|\) and \(G(K/{\mathbb Q}) \subset S_5\text{,}\) we know that \(G(K/{\mathbb Q})\) contains a cycle of length \(5\text{.}\) By Lemma 11.2.1, \(S_5\) is generated by a transposition and an element of order \(5\text{;}\) therefore, \(G(K/{\mathbb Q} )\) must be all of \(S_5\text{.}\) By Theorem 9.3.4, \(S_5\) is not solvable. Consequently, \(f(x)\) cannot be solved by radicals.
