- Show that every finite \(p\)-group is nilpotent.
Solution
Let \(G\) be a finite \(p\)-group. We prove that its upper central series reaches \(G\).
1. Every nontrivial quotient of \(G\) has nontrivial center.
If \(N\trianglelefteq G\) and \(G/N\ne1\), then \(G/N\) is again a finite \(p\)-group. Every finite \(p\)-group has nontrivial center by the class equation. Hence \[ Z(G/N)\ne1. \]
2. Define the upper central series by\[ Z_0(G)=1, \qquad Z_{i+1}(G)/Z_i(G)=Z(G/Z_i(G)). \]As long as \(Z_i(G)\ne G\), one has\[ Z_i(G)<Z_{i+1}(G). \]
If \(Z_i(G)\ne G\), then the quotient \(G/Z_i(G)\) is a nontrivial finite \(p\)-group. By 1 its center is nontrivial. Therefore \[ Z_{i+1}(G)/Z_i(G)=Z(G/Z_i(G))\ne1, \] which is equivalent to the strict inclusion.
3. The upper central series reaches \(G\) after finitely many steps.
Every \(Z_i(G)\) is a subgroup of the finite group \(G\). By 2, until the series reaches \(G\) its order strictly increases. More precisely, each quotient \[ Z_{i+1}(G)/Z_i(G) \] is a nontrivial \(p\)-group, so its order is at least \(p\). If \(|G|=p^n\), there can therefore be at most \(n\) strict inclusions. Hence for some \(c\le n\), \[ Z_c(G)=G. \]
4. Therefore \(G\) is nilpotent.
A group is nilpotent precisely when its upper central series reaches the whole group after finitely many steps. This is exactly 3.