Show that if \(G\) is solvable, then \(G\) contains a nontrivial normal subroup.
- What does this mean on the Galois theory side?
Hint: consider the derived series.
Solution
Taken literally, every nontrivial group \(G\) contains a nontrivial normal subgroup, namely \(G\) itself. The solvability hypothesis becomes meaningful if one asks for a proper nontrivial normal subgroup.
If \(G\) is nonabelian and solvable, then \[ 1\ne [G,G]\ne G. \] Indeed, \([G,G]\ne1\) because \(G\) is nonabelian, while \([G,G]\ne G\) because otherwise the derived series would be stationary at \(G\) and could not reach \(1\). Since \([G,G]\) is characteristic, it is normal. Thus every nonabelian solvable group has a proper nontrivial normal subgroup.
For abelian solvable groups this stronger assertion can fail: \(C_p\) is simple. More generally, a nontrivial abelian group has a proper nontrivial subgroup unless it has prime order.
For a finite Galois extension \(L/F\) with solvable Galois group \(G\), a normal subgroup \(H\trianglelefteq G\) corresponds to an intermediate field \(L^H\) that is Galois over \(F\). A proper nontrivial normal subgroup gives a proper nontrivial Galois intermediate field. The derived series gives the usual tower whose successive Galois groups are abelian.