Define what a composition series is, and state what it means for a group to be simple, solvable, or nilpotent.
- How are the derived and lower/upper central series defined? What type(s) of the groups above does each series correspond to?
Solution
A nontrivial group is simple if its only normal subgroups are \(1\) and itself.
A composition series of a finite group \(G\) is a subnormal series \[G=G_0\triangleright G_1\triangleright\cdots\triangleright G_n=1\] in which each factor \(G_i/G_{i+1}\) is simple; equivalently, each \(G_{i+1}\) is a maximal proper normal subgroup of \(G_i\). By the Jordan–Hölder theorem, the multiset of composition factors, up to isomorphism, does not depend on the composition series.
The derived series is \(G^{(0)}=G\), \(G^{(i+1)}=[G^{(i)},G^{(i)}]\), and \(G\) is solvable if \(G^{(n)}=1\) for some \(n\). Each \(G^{(i)}/G^{(i+1)}\) is abelian; conversely, if \(G=H_0\trianglerighteq H_1\trianglerighteq\cdots\trianglerighteq H_r=1\) has abelian factors, induction gives \(G^{(i)}\le H_i\), so \(G^{(r)}=1\). Thus \(G\) is solvable exactly when it has a finite subnormal series with abelian factors.
The lower central series is \(\gamma_1(G)=G\), \(\gamma_{i+1}(G)=[\gamma_i(G),G]\), and the upper central series is \(Z_0(G)=1\), \(Z_{i+1}(G)/Z_i(G)=Z(G/Z_i(G))\). The group \(G\) is nilpotent if \(\gamma_{c+1}(G)=1\) for some \(c\); equivalently, \(Z_c(G)=G\) for some \(c\), and the least such \(c\) is the same for both series, the nilpotency class.
Thus composition series record the simple factors, the derived series detects solvability, and the lower and upper central series detect nilpotence. Every nilpotent group is solvable, since \(G^{(i)}\le\gamma_{i+1}(G)\); the converse fails for \(S_3\), which is solvable with trivial center.