Let \(f\in K[x]\) be a separable polynomial of degree \(5\), and let \(G\) be the Galois group of its splitting field over \(K\).
- Give a necessary and sufficient permutation-theoretic condition for \(3\mid |G|\).
- If \(f\) is irreducible, show that \[ 3\mid |G| \quad\Longleftrightarrow\quad G\cong A_5\ \text{or}\ S_5. \]
- Give an irreducible quintic over \(\QQ\) for which \(3\nmid |G|\).
Solution
Let \(\Omega\) be the set of five distinct roots of \(f\) in a splitting field. Then \(G\) acts faithfully on \(\Omega\), so we regard \(G\) as a subgroup of \(S_5\).
1. We have \(3\mid |G|\) if and only if \(G\) contains a \(3\)-cycle.
If \(3\mid |G|\), Cauchy’s theorem gives an element of order \(3\) in \(G\). An element of order \(3\) in \(S_5\) must have cycle type \((3)(1)(1)\), hence is a \(3\)-cycle.
Conversely, a \(3\)-cycle has order \(3\), so if \(G\) contains one then Lagrange’s theorem gives \(3\mid |G|\).
2. Assume now that \(f\) is irreducible. Then \(3\mid |G|\) if and only if \(A_5\leq G\leq S_5\).
Irreducibility and separability imply that \(G\) acts transitively on the five roots, hence \(5\mid |G|\). By Cauchy’s theorem, \(G\) contains an element of order \(5\), which in \(S_5\) is a \(5\)-cycle.
Suppose \(3\mid |G|\). By 1, \(G\) also contains a \(3\)-cycle. Both a \(5\)-cycle and a \(3\)-cycle are even, so the subgroup \[ H:=G\cap A_5 \] contains elements of orders \(5\) and \(3\). Therefore \(15\mid |H|\). Since \(H\leq A_5\) and \(|A_5|=60\), the only possible orders are \(15\), \(30\), or \(60\).
The case \(|H|=30\) is impossible because an index-\(2\) subgroup of \(A_5\) would be normal, whereas \(A_5\) is simple. The case \(|H|=15\) is also impossible: Sylow’s theorems would force the Sylow \(5\)- and Sylow \(3\)-subgroups of \(H\) both to be normal, making \(H\cong C_{15}\); but \(A_5\) has no element of order \(15\). Hence \(|H|=60\), so \(H=A_5\). Thus \[ A_5\leq G\leq S_5, \] and consequently \(G=A_5\) or \(G=S_5\).
The converse is immediate, since both \(|A_5|=60\) and \(|S_5|=120\) are divisible by \(3\).
3. The polynomial \(x^5-2\in\QQ[x]\) is an irreducible example with \(3\nmid |G|\).
Eisenstein’s criterion at \(2\) shows that \(x^5-2\) is irreducible. Let \[ \alpha=2^{1/5}, \qquad \zeta=\zeta_5. \] Its splitting field is \[ L=\QQ(\alpha,\zeta). \] We have \[ [\QQ(\alpha):\QQ]=5, \qquad [\QQ(\zeta):\QQ]=\varphi(5)=4. \] The intersection \(\QQ(\alpha)\cap\QQ(\zeta)\) has degree over \(\QQ\) dividing both \(5\) and \(4\), hence is \(\QQ\). Therefore \[ [L:\QQ]=5\cdot4=20. \] Since \(L/\QQ\) is the splitting field of a separable polynomial, it is Galois, so \[ |G|=[L:\QQ]=20. \] Thus \(3\nmid |G|\).