- Define solvability by radicals for a polynomial \(f(x) \in F[x]\) and state the Abel–Ruffini / Galois criterion.
- Prove that the alternating group \(A_5\) is a simple group.
- Explain why the symmetric group \(S_5\) is not solvable, and why there is no general radical formula for quintic polynomials.
Solution
A polynomial \(f\in F[x]\) is solvable by radicals if its splitting field is contained in a tower \[ F=K_0\subset K_1\subset\cdots\subset K_r \] in which each \(K_{i+1}=K_i(\alpha_i)\) with \(\alpha_i^{n_i}\in K_i\) for some \(n_i\ge2\) (after adjoining the necessary roots of unity in the standard formulation). Galois theory gives the criterion: \(f\) is solvable by radicals if and only if its Galois group is solvable.
We prove that \(A_5\) is simple. Its conjugacy classes have sizes \[ 1,\ 20,\ 15,\ 12,\ 12, \] corresponding respectively to the identity, the \(3\)-cycles, the double transpositions, and the two \(A_5\)-classes of \(5\)-cycles. A normal subgroup is a union of conjugacy classes containing the identity, and its order must divide \(60\). The sums obtained from a proper nonempty selection of the four nontrivial class sizes are \[ 13,16,21,25,28,33,36,40,45,48, \] none of which divides \(60\). Hence the only normal subgroups are \(\{e\}\) and \(A_5\), so \(A_5\) is simple.
Now \[ 1\triangleleft A_5\triangleleft S_5 \] has composition factors \(A_5\) and \(S_5/A_5\cong C_2\). Since \(A_5\) is nonabelian simple, \(S_5\) is not solvable. Therefore any quintic whose Galois group is \(S_5\) is not solvable by radicals. Since quintics with Galois group \(S_5\) exist, there can be no radical formula valid for all quintic polynomials.
1. A polynomial \(f(x) \in F[x]\) is solvable by radicals if its roots lie in a radical extension tower:\[F = K_0 \subseteq K_1 \subseteq K_2 \subseteq \cdots \subseteq K_m\]where each step \(K_{i+1} = K_i(\alpha_i)\) with \(\alpha_i^{n_i} \in K_i\) for some integer \(n_i \ge 2\).
2. Theorem (Galois): A polynomial \(f(x) \in F[x]\) is solvable by radicals if and only if its Galois group \(\operatorname{Gal}(f/F)\) is a solvable group.
2. Simplicity of \(A_5\):
2.1. The alternating group \(A_5\) has order \(|A_5| = 5!/2 = 60\).
2.2. We compute the conjugacy classes of \(A_5\) by cycle types:- Identity: \(1\) element (\(e\)).- 3-cycles: \(\binom{5}{3} \times 2 = 20\) elements.- Products of two disjoint 2-cycles: \(\frac{1}{2} \binom{5}{2}\binom{3}{2} = 15\) elements.- 5-cycles: In \(S_5\) there are \(4! = 24\) 5-cycles. In \(A_5\), the centralizer of a 5-cycle has order 5, so the class splits into two \(A_5\)-conjugacy classes of size \(\frac{60}{5} = 12\) each.
2.3. The sizes of all conjugacy classes in \(A_5\) are:\[\{1, 15, 20, 12, 12\}.\]
2.4. Any normal subgroup \(N \trianglelefteq A_5\) must be a union of conjugacy classes containing \(e\) (\(1 \in N\)), and by Lagrange’s Theorem, \(|N|\) must divide \(|A_5| = 60\).
2.5. The possible proper divisors of 60 are \(1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30\).
2.6. Testing all sums of conjugacy class sizes containing 1:- \(1 + 12 = 13 \nmid 60\).- \(1 + 15 = 16 \nmid 60\).- \(1 + 20 = 21 \nmid 60\).- \(1 + 12 + 12 = 25 \nmid 60\).- \(1 + 12 + 15 = 28 \nmid 60\).- \(1 + 12 + 20 = 33 \nmid 60\).- \(1 + 15 + 20 = 36 \nmid 60\).- \(1 + 12 + 12 + 15 = 40 \nmid 60\).- \(1 + 12 + 12 + 20 = 45 \nmid 60\).- \(1 + 12 + 15 + 20 = 48 \nmid 60\).- \(1 + 12 + 12 + 15 + 20 = 60 = |A_5|\).
2.7. Thus the only divisors obtainable as class sums are \(1\) and \(60\).
2.8. Therefore, the only normal subgroups of \(A_5\) are \(\{e\}\) and \(A_5\), proving \(A_5\) is simple.
3. Insolvability of \(S_5\) and General Quintics:
3.1. The composition series of \(S_5\) is:\[\{e\} \triangleleft A_5 \triangleleft S_5.\]
3.2. The composition factors are:- \(S_5/A_5 \cong \mathbb{Z}_2\) (abelian).- \(A_5/\{e\} \cong A_5\) (non-abelian simple group of order 60).
3.3. Since \(A_5\) is simple and non-abelian (\(|A_5| = 60\) is not prime), its composition series cannot be refined to abelian factors.
3.4. Therefore, \(S_5\) is not solvable.
3.5. By Galois Theory, any quintic polynomial over \(\mathbb{Q}\) with Galois group \(S_5\) (e.g. \(f(x) = x^5 - 4x - 2\)) cannot be solved by radicals (Abel-Ruffini Theorem).
4. Conclusion:\(A_5\) is simple via class size sums \(\{1, 12, 12, 15, 20\}\), making \(S_5\) non-solvable and preventing radical solutions for quintics. Q.E.D.