Show that every finite \(p\)-group is solvable.
Solution
We argue by induction on \(|G|\).
If \(|G|=1\), the result is trivial. Let \(G\) be a nontrivial finite \(p\)-group. Its center is nontrivial, so choose a subgroup \[ C\le Z(G) \] of order \(p\). Then \(C\trianglelefteq G\) and \(C\) is abelian.
The quotient \(G/C\) is again a finite \(p\)-group of smaller order. By induction, \(G/C\) is solvable.
A group extension of a solvable group by a solvable normal subgroup is solvable. Since \(C\) is abelian, hence solvable, and \(G/C\) is solvable, it follows that \(G\) is solvable.
Equivalently, iterating this argument gives a normal series whose successive factors all have order \(p\), hence are cyclic and abelian.