Past exam of the number theory course of the University of Oxford / 2009 / A / 3

A finite group is solvable if its derived series, defined by $G^{(0)}=G$ and $G^{(j+1)}=[G^{(j)},G^{(j)}]$, reaches the identity after finitely many steps. Equivalently it has a subnormal series with abelian factor groups. The factors need not all have prime order for this definition.

Let $V_4=\{1,(12)(34),(13)(24),(14)(23)\}$. This is an abelian subgroup, and conjugation preserves its nonidentity cycle type, so it is normal in $S_4$ and hence in $A_4$. The serieshas factors $V_4\cong C_2\times C_2$, $A_4/V_4\cong C_3$, and $S_4/A_4\cong C_2$, all abelian. Thus $\boxed{S_4\text{ is solvable}}$.

The rotations form a cyclic subgroup $R\cong C_n$ of index two, so it is normal. The quotient $D_{2n}/R$ has order two and is cyclic. Hence $1⊴R⊴D_{2n}$ has abelian factors, proving $\boxed{D_{2n}\text{ is solvable}}$. Its commutator subgroup is contained in the abelian rotation subgroup, so its second derived subgroup is already trivial.

Let $\alpha =p^{1/p}$ and let $\zeta$ be a primitive $p$th root of unity. The roots are $\alpha {\zeta }^j$, so $L=\mathbb{Q}(\alpha ,\zeta )$. Eisenstein gives $[\mathbb{Q}(\alpha ):\mathbb{Q}]=p$, while the prime cyclotomic extension has degree $[\mathbb{Q}(\zeta ):\mathbb{Q}]=p-1$. Both degrees divide $[L:\mathbb{Q}]$, and they are coprime. Conversely adjoining $\alpha$ to the cyclotomic field has degree at most $p$. These two bounds giveRestriction to $E=\mathbb{Q}(\zeta )$ gives a normal subgroup $N=\mathrm{Gal}(L/E)$ of order $p$, hence cyclic, with quotient $\mathrm{Gal}(E/\mathbb{Q})\cong C_{p-1}$. Restriction is onto because embeddings of the intermediate field extend in the finite Galois splitting field. Thus $1⊴N⊴G$ has abelian factors, proving $\boxed{G\text{ is solvable}}$. For $p=2$ the cyclotomic field is just $\mathbb{Q}$, and the same argument gives degree two and a cyclic group.