Definitions and Basics
A normal series of a group \(G\) is a sequence \(G \to G^1 \to G^2 \to \cdots\) such that \(G^{i+1} {~\trianglelefteq~}G_i\) for every \(i\).
A central series for a group \(G\) is a terminating normal series \(G \to G^1 \to \cdots \to \left\{{e}\right\}\) such that each quotient is central, i.e. \([G, G^i] \leq G^{i-1}\) for all \(i\).
A composition series of a group \(G\) is a finite normal series such that \(G^{i+1}\) is a maximal proper normal subgroup of \(G^i\).
Any two composition series of a group have the same length and isomorphic composition factors (up to permutation).
A group \(G\) is simple iff \(H{~\trianglelefteq~}G \implies H = \left\{{e}\right\}, G\), i.e. it has no non-trivial proper subgroups.
If \(G\) is not simple, then \(G\) is an extension of any of its normal subgroups. I.e. for any \(N{~\trianglelefteq~}G\), \(G \cong E\) for some extension of the form \(N\to E\to G/N\).
Set \(G^0 = G\) and \(G^{i+1} = [G, G^i]\), then \(G^0 \geq G^1 \geq \cdots\) is the lower central series of \(G\).
Mnemonic: “lower” because the chain is descending. Iterate the adjoint map \([{-}, G]\), if this terminates then the map is nilpotent, so call \(G\) nilpotent!
Set \(Z_0 = 1\), \(Z_1 = Z(G)\), and \(Z_{i+1} \leq G\) to be the subgroup satisfying \(Z_{i+1}/Z_i = Z(G/Z_i)\). Then \(Z_0 \leq Z_1 \leq \cdots\) is the upper central series of \(G\).
Equivalently, since \(Z_i{~\trianglelefteq~}G\), there is a quotient map \(\pi:G\to G/Z_i\), so define \(Z_{i+1} \coloneqq\pi^{-1}(Z(G/Z_i))\) (?).
Mnemonic: “upper” because the chain is ascending. “Take higher centers”.
Set \(G^{(0)} = G\) and \(G^{(i+1)} = [G^{(i)}, G^{(i)}]\), then \(G^{(0)} \geq G^{(1)} \geq \cdots\) is the derived series of \(G\).
Solvability
A useful way to extract normal subgroups: let \(G\) act on literally anything by \(\phi: G\to \mathop{\mathrm{Aut}}(X)\). Then \(\ker \phi {~\trianglelefteq~}G\) is always a normal subgroup.
- \(G\curvearrowright G\) by \(x\mapsto gx\).
- \(G\curvearrowright\left\{{H\leq G}\right\}\) by \(H\mapsto gH\) or \(H\mapsto gHg^{-1}\).
- \(G\curvearrowright\left\{{{\operatorname{Syl}}_p(G)}\right\}\) for a fixed \(p\) by \(S_p \mapsto gS_p g^{-1}\).
- \(G\curvearrowright H\) for \(H{~\trianglelefteq~}G\) by inner automorphisms \(h\mapsto ghg^{-1}\).
A group \(G\) is solvable iff \(G\) has a terminating normal series with abelian composition factors, i.e. \begin{align*} G \coloneqq G_n > G_{n-1} > \cdots > G_2 > G_1 \coloneqq\left\{{ e }\right\} && \text{ with } G^{i}/G^{i+1}\text{ abelian for all } i .\end{align*}
If \(G = { \mathsf{Gal}}(L/K)\) is a Galois group corresponding to a polynomial \(f\), then \(G\) is solvable as a group iff \(f\) is solvable in radicals: there is a tower of extensions \(K = F_0 \subset F_1 \subset F_2 \subset \cdots \subset F_m = L\) where
- \(F_i = F_{i-1}(\alpha_i)\) where \(\alpha_i^{m_i } \in F_{i-1}\) for some power \(m_i \in {\mathbf{Z}}^{\geq 0}\), and
- \(F_m \supseteq \operatorname{SF}(f)\) contains a splitting field for \(f\).
A group \(G\) is solvable iff its derived series terminates.
If \(n\geq 4\) then \(S_n\) is solvable.
Some useful facts about solvable groups:
- \(G\) is solvable iff \(G\) has a terminating derived series.
- Solvable groups satisfy the 2 out of 3 property
- Abelian \(\implies\) solvable
- Every group of order less than 60 is solvable.