Discuss sufficient conditions on a polynomial of degree 5 to have Galois group \(S_5\) over \(\QQ\) and prove your statements.
Solution
Let \(f\in\mathbb Q[x]\) be irreducible of degree \(5\), and let \(G\) be its Galois group acting on the five roots. Irreducibility makes this action transitive.
If \(f\) has exactly three real roots and one nonreal conjugate pair, complex conjugation fixes the three real roots and swaps the two nonreal roots. Thus \(G\) contains a transposition.
A transitive subgroup \(G\le S_5\) containing a transposition equals \(S_5\): if \(\tau=(a\ b)\in G\), then all conjugates \(g\tau g^{-1}=(g(a)\ g(b))\) lie in \(G\). Form the graph on the five roots whose edges are these conjugate transpositions. The graph is \(G\)-invariant and contains an edge; transitivity forces it to be connected. Transpositions along the edges of a connected graph generate the full symmetric group, so \(S_5\le G\).
Hence a sufficient condition is: \[ \boxed{f\text{ irreducible of degree }5\text{ and having exactly three real roots}.} \] Equivalently, it is enough that the transitive Galois group contain a transposition. Irreducibility also implies \(5\mid |G|\), hence by Cauchy’s theorem \(G\) contains an element of order \(5\), necessarily a \(5\)-cycle.