Given a group \(G\), state the criteria that make \(G\) a solvable group. Use your definition to prove that \(S_4\) is solvable.
Let \(g(x) \in \mathbb{Q}[x]\) have degree 4. Prove or disprove: \(g(x)\) is not solvable by radicals.
Let \(g(x) \in \mathbb{Q}[x]\) have degree 5. Prove or disprove: \(g(x)\) is not solvable by radicals.
Solution
1. A group is solvable when its derived series reaches thetrivial subgroup after finitely many steps. Equivalently, it hasa finite subnormal series with abelian factors.
Define \(G^{(0)}=G\) and let \(G^{(j+1)}\) be the subgroup generated by all commutators \(xyx^{-1}y^{-1}\) with \(x,y\in G^{(j)}\). Solvability means \(G^{(n)}=1\) for some finite \(n\). Conjugation sends a commutator to the commutator of the conjugated elements, so each commutator subgroup is normal in its parent. The quotient by it is abelian because every commutator becomes trivial. Thus a terminating derived series is a finite series with abelian factors.
Conversely, suppose \(G=G_0\triangleright G_1\triangleright\cdots\triangleright G_n=1\) with every \(G_j/G_{j+1}\) abelian. Then \([G_j,G_j]\subseteq G_{j+1}\). Induction gives \(G^{(j)}\subseteq G_j\): taking commutators preserves the inclusion at each step. Therefore \(G^{(n)}\subseteq G_n=1\), so \(G\) is solvable under the derived-series definition.
2. The group \(S_4\) is solvable.
Let \[ V=\{1,(12)(34),(13)(24),(14)(23)\}. \] Each of its three nonidentity elements is an involution, and the product of two distinct ones is the third in either order. Hence \(V\) is an abelian subgroup of order \(4\). Conjugation in \(S_4\) permutes these double transpositions, so \(V\lhd S_4\). Also \(A_4\lhd S_4\), as the kernel of the sign map, and \(V\subseteq A_4\). The series \[ S_4\triangleright A_4\triangleright V\triangleright1 \] has factors of orders \(2\), \(3\), and \(4\). The first two are cyclic because their orders are prime, and the last is the abelian group \(V\cong C_2\times C_2\). All factors are abelian, so step 1 proves solvability.
3. The assertion in part (b) is false: every degree-fourpolynomial over \(\mathbb Q\) is solvable by radicals.
Let \(K\) be its splitting field. This is a finite Galois extension because the base field has characteristic zero. Its Galois group acts faithfully on the distinct roots: an automorphism fixing all roots fixes their generated field \(K\). There are at most four distinct roots, so labeling them and fixing any unused labels embeds \(\operatorname{Gal}(K/\mathbb Q)\) in \(S_4\). This argument includes reducible polynomials and repeated roots.
A subgroup \(H\) of a solvable group \(G\) is solvable because \(H^{(j)}\subseteq G^{(j)}\) for every \(j\), by induction on commutators. Step 2 therefore makes the Galois group solvable. The Galois criterion for solvability by radicals in characteristic zero states that a polynomial is solvable by radicals exactly when its splitting-field Galois group is solvable [@DF04]. Applying it proves the claim and disproves the proposed negation.
4. The assertion in part (c) is also false. An irreduciblecounterexample is \(g(x)=x^5-2\).
This polynomial is irreducible over \(\mathbb Q\) by Eisenstein’s criterion at \(2\) [@DF04]. Let \(a=\sqrt[5]{2}\), and let \(\zeta\) be a primitive fifth root of unity. Its five roots are \(a,\zeta a,\zeta^2a,\zeta^3a,\zeta^4a\).
The number \(\zeta\) is a root of the quartic \(x^4+x^3+x^2+x+1\). By step 3, all roots of that quartic lie in a radical extension of \(\mathbb Q\). Choose a radical tower ending in a field \(M\) containing them. Adjoining \(a\) is one further radical adjunction, since \(a^5=2\in M\). The resulting field \(M(a)\) contains all five displayed roots of \(g\). Thus \(g\) is solvable by radicals.