Show that every \(p\dash\)group is solvable.
Hint
Induct on \(\abs G\): \(Z(G)\) is nontrivial and abelian, \(G/Z(G)\) is a smaller \(p\)-group, and if \(N\normal G\) with \(N\) and \(G/N\) solvable then \(G\) is solvable.
Solution
1. Argue by induction on \(\abs G = p^n\).
2. Base case: if \(n \leq 1\) then \(G\) is trivial or of prime order, hence abelian, hence solvable.
3. Inductive step: let \(n \geq 2\) and assume every \(p\)-group of order less than \(p^n\) is solvable.
3.1. The class equation \[\abs G = \abs{Z(G)} + \sum_i [G : C_G(x_i)]\] over representatives of the conjugacy classes of size greater than \(1\) has every index divisible by \(p\), so \(p\) divides \(\abs{Z(G)}\) and \(Z(G) \neq 1\).
3.2. \(Z(G) \normal G\), and \(Z(G)\) is abelian, hence solvable.
3.3. \(\abs{G/Z(G)} = \abs G / \abs{Z(G)} < p^n\), and it is a power of \(p\), so \(G/Z(G)\) is solvable by the inductive hypothesis.
3.4. If \(N \normal G\) with \(N\) and \(G/N\) solvable, then \(G\) is solvable: lifting a subnormal series of \(G/N\) with abelian quotients through \(G \to G/N\) and appending a subnormal series of \(N\) produces one for \(G\).
3.5. Applying step 4 with \(N = Z(G)\) makes \(G\) solvable.
4. Q.E.D.
Steps 2 and 3 complete the induction, so every \(p\)-group is solvable.