Show that p-groups \(\implies\) nilpotent
Hint
A nontrivial finite \(p\)-group has nontrivial center, so each term \(Z_m\neq G\) of the upper central series is strictly smaller than \(Z_{m+1}\).
Solution
1. A nontrivial finite \(p\)-group has nontrivial center.
1.1. Let \(P\) be a group of order \(p^n\) with \(n \geq 1\), and take the class equation \[\abs P = \abs{Z(P)} + \sum_i [P : C_P(x_i)]\] over representatives \(x_i\) of the conjugacy classes of size greater than \(1\).
1.2. Each such index divides \(\abs P\) and exceeds \(1\), so \(p\) divides it.
1.3. \(p\) divides \(\abs P\), so \(p\) divides \(\abs{Z(P)}\), and \(Z(P) \neq 1\).
2. Let \(G\) have order \(p^n\) and let \(1 = Z_0 \leq Z_1 \leq \cdots\) be its upper central series, where \(Z_{m+1}\) is the preimage in \(G\) of \(Z(G/Z_m)\).
3. If \(Z_m \neq G\) then \(Z_{m+1} \neq Z_m\).
3.1. \(G/Z_m\) is a nontrivial group whose order divides \(p^n\), so it is a nontrivial \(p\)-group.
3.2. By step 1, \(Z(G/Z_m) \neq 1\).
3.3. \(Z_{m+1}/Z_m = Z(G/Z_m)\), so \(Z_{m+1}\) strictly contains \(Z_m\).
4. Q.E.D.
By step 3 the orders \(\abs{Z_0} < \abs{Z_1} < \cdots\) strictly increase while \(Z_m \neq G\), and they are bounded by \(p^n\). So \(Z_m = G\) for some \(m \leq n\), which is the definition of \(G\) being nilpotent.